inequalities
algebra
function
domain
algebra unsolved

Let $ x_1,x_2,...,x_5$ be nonnegative numbers such that $ x_1\plus{}x_2\plus{}x_3\plus{}x_4\plus{}x_5\equal{}5$.

Determine the maximum value of $ x_1x_2\plus{}x_2x_3\plus{}x_3x_4\plus{}x_4x_5$.

hello, i assume that $ x_1x_2\plus{}x_2x_3\plus{}x_3x_4\plus{}x_4x_5\le\frac{25}{4}$  and the
maximum will be attained at the point $ x_1\equal{}0,x_2\equal{}\frac{5}{8},x_3\equal{}\frac{5}{2},x_4\equal{}\frac{15}{8},x_5\equal{}0$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, i assume that $ x_1x_2 \plus{} x_2x_3 \plus{} x_3x_4 \plus{} x_4x_5\le\frac {25}{4}$  and the
maximum will be attained at the point $ x_1 \equal{} 0,x_2 \equal{} \frac {5}{8},x_3 \equal{} \frac {5}{2},x_4 \equal{} \frac {15}{8},x_5 \equal{} 0$.
Sonnhard.[/quote]
hello,
What do you mean by "i assume"?
You should always prove the answer you give.
Do you use some computer?
Else you wouldn't have put that complicated equality case where a simple one as $ (0,\frac{5}{4},\frac{5}{2},\frac{5}{4},0)$ exists. Actually there are infinitely many equality cases. Lets see that now.

Let $ f(x_1,x_2,x_3,x_4,x_5)\equal{}x_1x_2\plus{}x_2x_3\plus{}x_3x_4\plus{}x_4x_5$
$ f(0,x_1,x_2,x_3\plus{}x_1,x_4,x_5)\geq f(x_1,x_2,x_3,x_4,x_5)\iff x_1x_4\geq 0$ which is obviously true.
So maximum is achieved when $ x_1\equal{}0$. Similarly it can be proved that maximum is achieved when $ x_5\equal{}0$
So we need to find $ \max(x_3(x_2\plus{}x_4))$ when $ x_2\plus{}x_3\plus{}x_4\equal{}5$
$ x_3(x_2\plus{}x_4)\equal{}(\sqrt{x_3(x_2\plus{}x_4)})^2\leq (\frac{x_3\plus{}(x_2\plus{}x_4)}{2})^2\equal{}\frac{25}{4}$
Where the inequality step is $ AM\minus{}GM$.
So the required maximum is $ \boxed{\frac{25}{4}\text{ with equality at }\{x_1\equal{}x_5\equal{}0,x_3\equal{}x_2\plus{}x_4\equal{}\frac{5}{2}\}}$

Could someone explain why the method of Lagrange Multipliers gives a lesser max of $ \frac{25}{9}$?

[quote="skywalkerJ.L."]$ f(0,x_1,x_2,x_3 \plus{} x_1,x_4,x_5)\ge f(x_1,x_2,x_3,x_4,x_5)$

So the maximum is achieved when $ f(0,x_1,x_2,x_3 \plus{} x_1,x_4,x_5) \equal{} f(x_1,x_2,x_3,x_4,x_5)$ which leads to $ x_1 \equal{} 0$

Is it right?[/quote]
Yes. That's what I meant.
Edit: But I just noted now that $ x_4 \equal{} 0$ also gives that equality. Then we can prove that $ x_5 \equal{} 0$ when maximum is achieved and proceed in the same way to get the same maximum but some more equality cases:
$ \{x_2 \equal{} x_1 \plus{} x_3 \equal{}\frac {5}{2},x_4 \equal{} x_5 \equal{} 0\}\text{ and }\{x_4 \equal{}x_3 \plus{} x_5 \equal{}  \frac {5}{2},x_1 \equal{} x_2 \equal{} 0\}$
[quote="hsbhatt"]Could someone explain why the method of Lagrange Multipliers gives a lesser max of $ \frac {25}{9}$?[/quote]
I don't know anything about Lagrange Multipliers. But For $ (\frac {5}{2},\frac {5}{2},0,0,0)$ gives the value $ \frac {25}{4}$. So even if my solution is wrong, $ \max\geq \frac {25}{4}$.
Maybe you could post the solution using Lagrange Multipliers (if possible). So it would be easier for someone to point out the problem.

The query was addressed to someone who knew Lagrange Multipliers as it is pretty straightforward calculation. A second check gave me a more ludcirous answer

I just wanted to know why Lag Mult does not give the max in this case?

I think because Langrange multiplier theorem works on an open domain.
So you need to take x_i in the interval [0,5]. You know that there is a maximum and a minimum because the domain is compact.
 Since there are no extremums in the product of intervals (0,5), you have to check separately for each instance x_i=0.


Can $\sin{x}* \cos{x}$ be expressed in another way? Or $\sin{ax}* \cos{ax}$ when a is a constant not equal to zero?

[hide="Well..."] $\sin ax \cos ax = \frac{1}{2}\sin 2ax$ [/hide]

$sin ax cos ax$ can equal a number of things.
$sin 2x = 2sin x cos x$ implying what was said.
Or it can be described by the $sin ax$ and $cos ax$ individualy by using the angle addition formulae of each; it would just need to be iterated.
The sum of angles for each is:
$sin(\alpha+\beta)= sin\alpha cos\beta+cos\alpha sin\beta$
$cos(\alpha+\beta) = cos\alpha cos\beta-sin\alpha sin\beta$.
All you would have to do it substitute any $\alpha$ or $\beta$, within some reason of course LOL  :wink: , in this case you might want a multiple of $x$.

EDIT:  :huh: The formula for angle addition in cosine is with a minus. opps ahahahah

I thought

$\sin (\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta$?

[quote="7h3.D3m0n.117"]I thought

$\sin (\alpha \pm \beta) = \sin \alpha \cos \beta \mp \cos \alpha \sin \beta$?[/quote]

It's wrong!

Yeah I have to flip the second plus or minus sign  :D . But other than that it's right.


trigonometry
geometry
trig identities
Law of Sines
geometry unsolved

Points $ D,E$ lie on side $ AB$ of $ \triangle{ABC}$ and satisfy $ \frac{AD}{DB}\cdot \frac{AE}{EB}\equal{}(\frac{AC}{CB})^2$.Prove that $ \angle{ACD}\equal{}\angle{BCE}$

[hide="solution"]

Consider D' such that m<ACD' and m<BCE are equal.

Applying Law of sines to triangle ACD', we get $ \frac{AD'}{\sin{ACD'}} \equal{} \frac{AC}{\sin{AD'C}}$

Also apply it for trianlge BCD' to get $ \frac{D'B}{\sin{BCD'}} \equal{} \frac{BC}{\sin{180\minus{}AD'C}}$

Dividing, we get $ \frac{BD'}{D'A} \equal{} \frac{BC}{AC} * \frac{\sin{BCD'}}{\sin{D'CA}}$

We do similarly for E. Multiplying, we get that $ \frac{AD'}{D'B}\cdot \frac{AE}{EB}\equal{}(\frac{AC}{CB})^2.$
Since $ \frac{AD}{DB}\cdot \frac{AE}{EB}\equal{}(\frac{AC}{CB})^2$,

We have $ \frac{AD}{DB} \equal{} \frac{AD'}{D'B}$

Which implies D=D'.

[/hide]

[quote="anonymous111"]Points $ D,E$ lie on side $ AB$ of $ \triangle{ABC}$ and satisfy $ \frac {AD}{DB}\cdot \frac {AE}{EB} \equal{} (\frac {AC}{CB})^2$.Prove that $ \angle{ACD} \equal{} \angle{BCE}$[/quote]
[hide="Solution"]
Consider $ X$ and $ Y$ on $ BC$ and $ AC$ respectively so that $ BC\parallel DY$ and $ EX\parallel AC$.  Then, $ \frac{EX}{AC}\equal{}\frac{BE}{AB}$.  Also, $ \frac{AD}{AB}\equal{}\frac{DY}{BC}$.  Then, $ \frac{EX}{DY}\cdot \frac{BC}{AC}\equal{}\frac{BE}{AD}$, so $ \frac{EX}{DY}\equal{}\frac{BE}{AD}\cdot \frac{AC}{BC}$.  Furthermore, $ \frac{XC}{BC}\equal{}\frac{EA}{AB}$ and $ \frac{CY}{AC}\equal{}\frac{BD}{AB}$, so $ \frac{XC}{CY}\cdot \frac{AC}{BC}\equal{}\frac{EA}{BD}$.  Then, $ \frac{CY}{XC}\equal{}\frac{AC}{BC}\cdot \frac{BD}{EA}$.  It follows that \[ \frac{CY}{XC}\cdot \frac{EX}{DY}\equal{}\frac{AC^2}{BC^2}\cdot \frac{BD}{EA}\cdot \frac{BE}{AD}\equal{}1\implies \frac{CY}{DY}\equal{}\frac{XC}{EX}\] This, combined with the fact that $ \angle CXE\equal{}180\minus{}\angle ACB\equal{}\angle CYD$ implies that $ \triangle CXE\sim \triangle CYD$, so $ \angle BCE\equal{}\angle ACD$. [/hide]

Dear Mathlinkers,
this relation is well knowed.
Se for example  R. Johnson Advanced Euclidean Geometry p. 244.
Sincerely
Jean-Louis

You can prove it by $ S_\triangle$ or $ T(A,k,l)$

[hide]Let us to restate the point $ D,\ E,$ on the side-segment $ BC$ and satisfy $ \frac {DB}{DC}\cdot \frac {EB}{EC} = \left(\frac {AB}{AC}\right)^{2}$ $ ,(1).$ We will prove that $ \angle BAD \ = \angle CAE.$

We suppose that $ \angle BAD\neq \angle CAE$ and then, let $ D'$ be, the point on $ BC$ such that $ \angle BAD' = \angle CAE$ $ ,(2)$

From $ (2)$ we conclude that $ \frac {D'B}{D'C}\cdot \frac {EB}{EC} = \left( \frac {AB}{AC}\right)^{2}$ $ ,(3)$ as well known and easy to prove.

From $ (1),$ $ (3)$ $ \Longrightarrow$ $ \frac {D'B}{D'C} = \frac {DB}{DC}$ $ \Longrightarrow$ $ D'\equiv D$ and we are done.[/hide]
Kostas Vittas


Identify A to E  :maybe:

Napthaline + Pthalic anhydride + $ AlCl_3$ +heat ------> A

A + $ Hydrazine/OH^{\minus{}}$ and heat ----> B

B  + $ PCl_5$ followed by $ AlCl_3$ and heat -----> C

C + $ CH_3MgI$ followed by hydrolysis -------> D

D + acidification and then heat with Se ---------> E

[hide]Please give mechanism for first step if possible. Seems friedel-crafts........but I'm not able to write the product  :( [/hide]

The orientation of Friedel-Crafts acylations of naphthalenes are dependent of the solvent, which you don't provide. Therefore, in the following I will assume beta-substitution.

[hide="Compound E"]5-Methylnaphthacene.[/hide]

Shouldn't the attack of the electrophile be on the 5 or 8 positions  :?:  They are more reactive sites , right? I'm not sure :blush: .

What 5 or 8 positions? In an unsubstituted naphthalene ring there are only two positions, 1 and 2 (or alpha and beta).

As I said before, the position of attack depends on the solvent used.

Any references or links as to the type of solvent needed?

No, but I will sumarize it for you:

The effect of solvent is felt mainly in Friedel-Crafts acylations, in which the solvent carbon dissulfide gives mainly alpha substitution, and the solvent nitrobenzene gives mainly beta substitution.

But the solvent is not mentioned , so how did you proceed?

Check what I wrote above:

[quote="Carcul"]Therefore, in the following I will [b][i]assume[/i][/b] beta-substitution.[/quote]

Isn't postion 1 more reactive than position 2?

One more time, that depends on the conditions. For example, if the attacking species is too bulky, than position two is more reactive, since in position 1 there is steric hindrance with hydrogen in carbon 8.


[quote="2006"]$\ a,b,c\ghe0$ .
$\frac{a}{a+\sqrt{(a+b)(a+c)}}+\frac{b}{b+\sqrt{(b+c)(b+a)}}+\frac{c}{c+\sqrt{(c+a)(c+b)}}\leq1$
     Good luck  :D[/quote]
$\frac{a}{a+\sqrt{(a+b)(a+c)}}\leq\frac{\sqrt a}{\sqrt{a}+\sqrt{b}+\sqrt{c}}.$  :D


linear algebra
matrix
calculus
derivative
algebra
polynomial
calculus computations

$ f(x,y,z) \equal{} x^2 \minus{} 4y^2 \plus{} 2z^2 \minus{} 3xy \plus{} 4yz \plus{} 2y \minus{} 3z$. A stationary point is $ \left(\frac {5}{11}, \frac {10}{33}, \frac {59}{132} \right)$.

The Hessian is $ \bold{H}_{(x,y,z)} \equal{} \begin{bmatrix} 2 & \minus{} 3 & 0 \\
\minus{} 3 & \minus{} 8 & 4 \\
0 & 4 & 0 \end{bmatrix}$. Now I think the eigenvalues are complex (so can we conclude that the point is a saddle point?) The second derivative test applies only for real eigenvalues.

i.e. $ \lambda^{3} \plus{} 6 \lambda^{2} \minus{} 41 \lambda \plus{} 32 \equal{} 0$ for $ \lambda \in \mathbb{C}$.

It has three real roots actually.
One negative root, two positive roots.
You can check this by plugging in 0 and 3 for the polynomial.

Remark: the eigenvalues of a real symmetric matrix must be real.

[quote="ragingbullrocky"]you get 38 for $ \lambda \equal{} 0$.[/quote]
What I mean is, the polynomial is continuous. Being a degree-three polynomial, you know that $ f(x) \rightarrow \minus{} \infty$ as $ x \rightarrow \minus{} \infty$, and $ f(x) \rightarrow \infty$ as $ x \rightarrow \infty$. If the polynomial (let us call it $ g(x)$) satisfies $ g(0) \equal{} 32 > 0$ and $ g(3) \equal{} \minus{} 10 < 0$, you know that there is one root in $ ( \minus{} \infty, 0)$, one in $ (0,3)$, and one in $ (3,\infty)$.


The following problem was posted on the "Getting Started" page.

you have a string and you make a mark every seventh and every thirteenth of the string. Then you divide it into 20 equal pieces. Prove that every piece except for the first and last piece will have only one marking.

Since 7+13=20, I suggest the following generalization.

You have a string and you make a mark every mth and every nth of the string. Then you divide it into (m+n) equal pieces. Prove or disprove:  If m and n are relatively prime positive integers, every piece except for the first and last piece will have only one marking.

PS:  I don't know the answer or proof.  Thus I am only hoping that it belongs on the Intermediate, not the Advanced page.

Darn.. just as I was about to give the proof :)

This was my favorite problem on it...it has a neat solution which I will not elaborate on now, of course.

Thanks for catching that!

Pure coincidence  - I'm not even aware of that program/test.

After the due date, i would like to see those proofs.

[quote="s0mp"]This is similar to a problem on the current HCSSiM application.[/quote]

i have a proof of this problem in my algebra book.  Could i get in with that?

There are a few other problems, so no, but I'm sure you could solve the other ones.

[quote="Scrambled"]i have a proof of this problem in my algebra book.  Could i get in with that?[/quote]

You would also be obligated to tell them that you were getting the proof out of a book, or at least tell them that you had previously learned the proof.


function
calculus
integration
complex analysis
complex numbers
absolute value

If $ f\equal{}u\plus{}iv$ be analytic on the disk $ |z|\leq R$, then
$ f(z)\equal{}\frac{1}{2\pi i}\oint_{|\zeta|\equal{}R}\frac{\zeta\plus{}z}{\zeta\minus{}z}\frac{u(\zeta)}{\zeta}\ d\zeta\plus{}iv(0)$ for $ |z|<R$

Let $ f$ be an entire function whose real part satisfies $ |\Re f(z)|\leq M|z|^{2}$ for all sufficiently large values of $ |z|$, where $ M$ is constant.  Use the above formula with $ R\equal{}2|z|$ to show that $ |f(z)|$ is bounded for large $ |z|$ by some multiple of $ |z|^{2}$, and hence, $ f$ must be a polynomial.

For large $ |z|$ the expression $ \frac {\zeta \plus{} z}{\zeta \minus{} z}$ has absolute value close to $ 1$ (most of the time) and $ \frac {u(\zeta)}{\zeta}$ has absolute value bounded by $ \frac {M|z|^2}{R} < M|z|$.  Can you find an upper bound on the value of the contour integral?

Yes, I just recently solved it.
$ \left|\int f\right|\leq \ell(\Gamma)\cdot\max|f|$
where $ \ell(\Gamma)$ is the length of the contour.


Just make the usual substitution $ a=tgx, b=tgy, c=tgz$ and after that use Jensen inequality.

[quote="carlosbr"]For $a,b,c $ be positive real numbers,
	such that: $a+b+c = abc$.
Show that:
$\frac{1}{\sqrt{1+a^2}} + \frac{1}{\sqrt{1+b^2}} +\frac{1}{\sqrt{1+c^2}} \leq \frac{3}{2},$ and sohow when the equality ocurrs.[/quote]

Putting $a=1/A, b=1/B,c=1/C$ we have $AB+AC+BC=1$ and we need to show that
\[\frac{A}{\sqrt{1+A^2}} + \frac{B}{\sqrt{1+B^2}} +\frac{C}{\sqrt{1+C^2}} \leq \frac{3}{2}\].

But $AB+AC+BC=1$, and so we need to show that:

\[\frac{A}{\sqrt{AB+AC+BC+A^2}} + \frac{B}{\sqrt{AB+AC+BC+B^2}} +\frac{C}{\sqrt{AB+AC+BC+C^2}} \leq \frac{3}{2}\].

We will prove this inequality without the restriction $AB+AC+BC=1$, because it is homogeneous.

Look at this: $AB+AC+BC+A^2=A(A+B+C)+BC$.
We will suppose that $A+B+C=1$ now.
So, $A(A+B+C)+BC=A+BC=(1-B-C)+BC=(1-B)(1-C)=(A+C)(B+C)$.

We just need to show that
\[\frac{A}{\sqrt{(1-B)(1-C)}} + \frac{B}{\sqrt{(1-A)(1-C)}} +\frac{C}{\sqrt{(1-A)(1-B)}} \leq \frac{3}{2}\].

Putting $1-A=x^2, 1-B=y^2, 1-C=z^2$, we can write
$1-x^2+1-y^2+1-z^2=1 \Leftrightarrow x^2+y^2+z^2=2$ and we need to show that
\[\frac{1-x^2}{yz} + \frac{1-y^2}{xz} +\frac{1-z^2}{xy} \leq \frac{3}{2}\].
Opening this expression:
\[\frac{x-x^3+y-y^3+z-z^3}{xyz} \leq \frac{3}{2}\].
\[2({x-x^3+y-y^3+z-z^3}) \leq 3{xyz}\].
\[2((x+y+z)-(x^3+y^3+z^3)) \leq 3{xyz}\].
\[(x+y+z)(x^2+y^2+z^2) \leq 2(x^3+y^3+z^3) + 3{xyz}\].

It is just another form of Schur Inequality!


Please give me hints only for the following problem. Thank you!

A quadrilateral inscribed in a circle has side lengths whose squares are 17, 99, 19, and 97. Find the area of the circle.

[size=75]Source: Mandelbrot Sample Regional Test[/size]

Brahmagupta's Formula:

[url]http://en.wikipedia.org/wiki/Brahmagupta%27s_formula[/url]

EDIT: oh oops...

thought we were to find the area of the quadrilateral...

Is there an easier way? I tried that but it got really messy.

Brahmagupta's Formula doesn't work because you're finding the area of the circle, and not the area of the quadrilateral.

[hide="Any comments on this?"]

Seems quite vague. Is the area:  29 pi?

Assuming the sum of squares of two sides pair wise we can see that that the diagonal forms the common base, in fact the hypotenuse of a right angled triangle.

So the 99 + 17 or 97 + 19  would be the square of the third side i.e. 116... This would be the diameter of the circle and hence the area would be 29 pi.

Honestly i am not sure :)[/hide]

[hide="Hints"]
Use Ptolemy's Theorem to find the lengths of the diagonals.
[hide="Another Hint"]
You now have many triangles. What is the circumradius of one of these triangles?
[/hide]
[/hide]

EDIT: Major fail. Ignore above.

[hide="Big Hint"]Notice 17 + 99 = 19 + 97[/hide]

[hide="solution"]From above observation the diagnol must be the diameter of the circle. Thus the Diameter has length $ \sqrt {116} \implies \pi r^2 \equal{} \pi \times \frac {116}{4} \equal{} 29\pi$[/hide]

In response to math154's second hint:
the circumradius of a triangle $ ABC \equal{} \frac {abc}{4(ABC)}$
Where (ABC) denotes the area

Ptolemy's theorem : $ ac \plus{} bd \equal{} ef$ (I dont see how that helps) :wink:
where a,b,c,d are the sides and e,f are the diagnols.


AMC
AIME
geometry
ratio
similar triangles

Number 3 on the 1984 AIME

P is a point inside the triangle ABC. Lines are drawn through P parallel to the sides of the triangle. The areas of the three resulting triangles with a vertex at P have areas 4, 9 and 49. What is the area of ABC?

[hide]

Notice similar triangles

Take a look at the ratio of sides between the triangles.  they are in ratio of 2:3:7

Now BC is made up of 3 segments, the side of the shaded triangle touching it and then the 2 other segments...which are the same size as the sides from the other shaded triangles.  So if the shaded triangle touching BC is of area 49, then BC = 1+2/7+3/7 =12/7 times the side of the side of the adjacent shaded triangle, so the area of ABC =(12/7)^2*49=144.

Just have to notice that a side of the big triangle is the sum of the similar sides of the little ones.

[/hide]

Quote:How did you figure the triangle sides were in the ratio of 2:3:7? Also, which triangle sides are you talking about exactly?

[hide]Since the lines are parallel you can prove that the triangles are similar. The ratio of the sides of similar triangles are equal.  The ratio of their areas is the ratio of their corresponding sides squared.  See if you can prove that if two triangles are similar the ratio of there areas is equal to the square of the ratio of corresponding sides.[/hide]


Let $ A_1$ and $ A_2$ be matrices of the same size and $ V_1$, $ V_2$ be the spaces spanned by the rows of $ A_1$ and $ A_2$ respectively. Prove that $ dim (V_1\plus{}V_2)\equal{} rank(A_1\plus{}A_2)$.

[img]http://upload.wikimedia.org/wikipedia/commons/1/12/O_RLY.jpg[/img]

  darij


Let $a$ and $b$ reals such that $ab>0$. Find all real functions $f$ for which
\[f(ax^2+bxy+f(y))=bf^2(x)+axy+y\] for all $x$ and $y$.

[i]Edited by Myth[/i]

I think, it is a time to write a solution, harazi!

If only I remembered the idea. I will try to solve it. Anyway, the guest seems to dislike it very much.  :)

It is really not easy. I have thought on it an for an hour and all that  I get is that $f^{2}(x)=\frac{a^{2}}{b^{2}}x^{2}+f^{2}(0).$
That's all for now.I am too tired now, I'll try to finish it tomorrow.     :)

I came, as the guest did, at the conclusion that the problem is really boring (it is not the first one I post, of course). Here is a short solution.  Taking x=0 we find that the function is surjective and that $ f(f(x))=bf^2(0)+x$. Take $u$ such that $f(u)=0$. Next, put $ x=-\frac{ax}{b}$. We find $ f(f(y))=bf^2(-\frac{by}{a})+y-by^2$. From these relations we get $ f^2(0)+y^2=f^2(-\frac{by}{a})$. Take now $ by+au=0$ and find that $ f(0)=0$ and $ f(f(x))=x$. Take $y=0$ in the original relation and find that $  f(ax^2)=bf^2(x)$. Combine this with $ f^2(-\frac{by}{a})=y^2$ and conclude using these two relations. From here it is just boring, not difficult.

Ok, guys, I have no time now to read[b] harazi[/b]'s   solution, so my solution can be very similiar to his one. Anyway here is it:Put $x=0$
 to get   $f(f(y))=bf^{2}(0)+y$ (*) for all $y$.Then put $y=-\frac{ax}{b}\Rightarrow f(f(-\frac{ax}{b}))=bf^{2} (x)-\frac{a^{2}x^{2}}{b}-\frac{ax}{b}$ (**).From (*)  and (**) we obtain: $f^{2}(x)=\frac{a^{2}}{b^{2}}x^{2}+f^{2}(0)$ (***)for all $x$. From (*) we get $f^{2}(f(y))=b^{2}f^{4}(0)+2byf^{2}(0)+y^{2}$, on the other hand from (***) we get that $f^{2}(f(y))=\frac{a^{4}}{b^{4}}y^{2}+\frac{a^{2}}{b^{2}}f^{2}(0)+f^{2}(0)$ for all $y$. So $y^{2}(\frac{a^{4}}{b^{4}}-1)-2byf^{2}(0)+f^{2}(0)-b^{2}f^{4}(0)=0$ for all y$\Rightarrow a=b$, and $f(0)=0$.From that we obtain: $f^{2}(x)=x^{2}$.If we just square up our original equality we get $f(y)=y$ for all $y$.
My solution is not long, I've just wrote all the calculations.
Is my solution same as [b]harazi[/b]'s one?

You have no time to read harazi's solution, but you have time to write your own.  :lol:
At the first sight, it is not the same as harazi's solution.

Actually, [b]Myth[/b], I have typed my solution this morning and only submitted it here in the evening.So I wasn't writing my solution online. :)


limit
function
logarithms
algebra unsolved
algebra

Define the following sequence as:
$ x_n \equal{} n(\sqrt [n]{a} \minus{} 1),(a > 0,a\neq 1)$.
1.Prove that sequence:
a)$ x_{2^k}$
b)$ x_n$ is decreasing.
2.Prove that in sequence exist a limit,$ l(a)$.
3.Prove that $ l(ab) \equal{} l(a)\plus{}l(b)$.

[quote="Erken"]Define the following sequence as:
$ x_n \equal{} n(\sqrt [n]{a} \minus{} 1),(a > 0,a\neq 1)$.
1.Prove that sequence:
a)$ x_{2^k}$
b)$ x_n$ is decreasing.
2.Prove that in sequence exist a limit,$ l(a)$.
3.Prove that $ l(ab) \equal{} l(a)l(b)$.[/quote]
What is a)?
b) followed from $ \frac {\partial x_y}{\partial y} < 0, y > 1$.
2. $ \lim_{y \to \infty}x_y \equal{} \lim_{z\to 0}\frac{e^z \minus{} 1}{z ln(a)} \equal{} \frac{1}{ln(a)}$, therefore 3. is not true.

Consider  the function:
$ f(x) = \frac {a^x - 1}{x}$ where $ x > 0$
$ f'(x) = \frac {a^x\ln{a} - a^x + 1}{x}$
We prove that $ f'(x) > 0\Leftrightarrow a^x\ln{a} - a^x + 1\geq 0$
Consider :
$ g(x) = xa^x\ln{a} - a^x + 1$
$ g'(x) = xa^x(\ln{a})^2 + a^x\ln{a} - a^x\ln{a} > 0$
And $ g(0) = 0$ so $ g(x) > 0,\forall x > 0$
$ \Rightarrow f'(x)\geq 0$
$ f(0) = 0$ so $ f(x)$ monotone on $ (0, + \infty)$
Let $ x = \frac {1}{n}$
Apply we has $ \{x_n\}$ is decrease.
3)when $ n\to\infty$ then $ x\to 0$
$ \lim_{x\to 0}f(x) = \lim_{x\to\0} {a^xlna} = ln{a}$
4)Must be $ \lim l(ab) = l(a) + l(b)$

[quote="TTsphn"]
4)Must be $ \lim l(ab) \equal{} l(a) \plus{} l(b)$[/quote]
Yes,you're right.Edited already.Sorry


[quote="ductrung"]Prove that for all positive real numbers $ a,b$ and $ c$,
\[ \frac {bc}{a^2 \plus{} bc} \plus{} \frac {ca}{b^2 \plus{} ca} \plus{} \frac {ab}{c^2 \plus{} ab}\le \frac {1}{2k \plus{} 1}\left( 3k \plus{} \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)
\]
where $ k \equal{} 1/2$.[/quote]
Turn out easy. :maybe:
$ \sum_{cyc}\frac {bc}{a^2 \plus{} bc}\leq\frac {1}{2}\left(\frac {3}{2} \plus{} \sum_{cyc}\frac {a}{b \plus{} c}\right)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(b \plus{} c \plus{} 2a)(a^2 \minus{} bc)}{(b \plus{} c)(a^2 \plus{} bc)} \plus{} \sum_{cyc}\frac {2bc(2a \minus{} b \minus{} c)}{(b \plus{} c)(a^2 \plus{} bc)}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {(b \plus{} c \plus{} 2a)(a \minus{} b)(a \plus{} c)}{(b \plus{} c)(a^2 \plus{} bc)} \minus{} \frac {(b \plus{} c \plus{} 2a)(c \minus{} a)(a \plus{} b)}{(b \plus{} c)(a^2 \plus{} bc)}\right) \plus{}$
$ \plus{} \sum_{cyc}\left(\frac {4bc(a \minus{} b)}{(b \plus{} c)(a^2 \plus{} bc)} \minus{} \frac {4bc(c \minus{} a)}{(b \plus{} c)(a^2 \plus{} bc)}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)\left(\frac {(b \plus{} c \plus{} 2a)(a \plus{} c)}{(b \plus{} c)(a^2 \plus{} bc)} \minus{} \frac {(a \plus{} c \plus{} 2b)(b \plus{} c)}{(a \plus{} c)(b^2 \plus{} ac)}\right) \plus{}$
$ \plus{} \sum_{cyc}(a \minus{} b)\left(\frac {4bc}{(b \plus{} c)(a^2 \plus{} bc)} \minus{} \frac {4ac}{(a \plus{} c)(b^2 \plus{} ac)}\right)\geq0\Leftrightarrow$
$ \sum_{cyc}\frac {(a \minus{} b)^2(c^4 \plus{} (a \plus{} b)c^3 \plus{} 2abc^2 \plus{} (2a^3 \minus{} 3a^2b \minus{} 3ab^2 \plus{} 2b^3)c \plus{} a^2b^2)}{(a \plus{} c)(b \plus{} c)(a^2 \plus{} bc)(b^2 \plus{} ac)}\geq0.$
But $ c^4 \plus{} (a \plus{} b)c^3 \plus{} 2abc^2 \plus{} (2a^3 \minus{} 3a^2b \minus{} 3ab^2 \plus{} 2b^3)c \plus{} a^2b^2\geq$
$ \geq2abc^2 \plus{} (2a^3 \minus{} 3a^2b \minus{} 3ab^2 \plus{} 2b^3)c \plus{} a^2b^2\geq$
$ \geq(2a^3 \minus{} 3a^2b \plus{} 2(ab)^{\frac {3}{2}} \minus{} 3ab^2 \plus{} 2b^3)c\geq0.$
There is a stronger inequality. :wink:

[quote="arqady"]
$ \sum_{cyc}\frac {(a \minus{} b)^2(c^4 \plus{} (a \plus{} b)c^3 \plus{} 2abc^2 \plus{} (2a^3 \minus{} 3a^2b \minus{} 3ab^2 \plus{} 2b^3)c \plus{} a^2b^2)}{(a \plus{} c)(b \plus{} c)(a^2 \plus{} bc)(b^2 \plus{} ac)}\geq0.$
[/quote]
Should be
$ \sum_{cyc}\frac {(a \minus{} b)^2(c^4 \plus{} 3(a \plus{} b)c^3 \plus{} 2abc^2 \plus{} (2a^3 \minus{} 3a^2b \minus{} 3ab^2 \plus{} 2b^3)c \plus{} a^2b^2)}{(a \plus{} c)(b \plus{} c)(a^2 \plus{} bc)(b^2 \plus{} ac)}\geq0.$
otherwise, the rest is OK. My solution is slightly simpler.

[quote="arqady"]There is a stronger inequality. :wink:[/quote]
Yes, arqady. That's why I presented the inequality in its original form. But I only have nice solution for $ 0\le k\le 1/2$.

Dear Ductrung, which step of Arqady's solution do you make simpler?

Thank you very much.

P.S. Arqady solution works, but in my opinion it is not simple to find. :wink:

[quote="ductrung"][quote="arqady"]
$ \sum_{cyc}\frac {(a \minus{} b)^2(c^4 \plus{} (a \plus{} b)c^3 \plus{} 2abc^2 \plus{} (2a^3 \minus{} 3a^2b \minus{} 3ab^2 \plus{} 2b^3)c \plus{} a^2b^2)}{(a \plus{} c)(b \plus{} c)(a^2 \plus{} bc)(b^2 \plus{} ac)}\geq0.$
[/quote]
Should be
$ \sum_{cyc}\frac {(a \minus{} b)^2(c^4 \plus{} 3(a \plus{} b)c^3 \plus{} 2abc^2 \plus{} (2a^3 \minus{} 3a^2b \minus{} 3ab^2 \plus{} 2b^3)c \plus{} a^2b^2)}{(a \plus{} c)(b \plus{} c)(a^2 \plus{} bc)(b^2 \plus{} ac)}\geq0.$
[/quote]
Yes, you are right:
$ (a \plus{} c)^2(b \plus{} c \plus{} 2a)(b^2 \plus{} ac) \minus{} (b \plus{} c)^2(a \plus{} c \plus{} 2b)(a^2 \plus{} bc) \equal{}$
$ \equal{} (a^2 \plus{} 2ac \plus{} c^2)(b^3 \plus{} b^2c \plus{} 2b^2a \plus{} abc \plus{} c^2a \plus{} 2a^2c) \minus{}$
$ \minus{} (b^2 \plus{} 2bc \plus{} c^2)(a^3 \plus{} a^2c \plus{} 2a^2b \plus{} abc \plus{} c^2b \plus{} 2b^2c) \equal{}$
$ \equal{} ... \plus{} (b^2 \plus{} ab \plus{} 2a^2 \plus{} 2a^2 \minus{} a^2 \minus{} ab \minus{} 2b^2 \minus{} 2b^2)c^3 \plus{} ... \equal{}$
$ \equal{} (a \minus{} b)(... \plus{} 3(a \plus{} b)c^3 \plus{} ...).$
$ 4bc(a \plus{} c)(b^2 \plus{} ac) \minus{} 4ac(b \plus{} c)(a^2 \plus{} bc) \equal{} (a \minus{} b)(... \plus{} 0c^3 ).$
But it does not change nothing: your inequality is true! :lol:

arqady's solution is totally valid - I just correct a typo, that's all.
My solution is as follows: rewrite the inequality as
$ \sum_{cycl}\left(\frac{2bc}{a^2\plus{}bc}\minus{}1\right)
\le\frac{1}{2k\plus{}1}\sum_{cycl}\left(\frac{2a}{b\plus{}c}\minus{}1\right)$
$ \Leftrightarrow \sum_{cycl}\frac{bc\minus{}a^2}{a^2\plus{}bc}
\le\frac{1}{2k\plus{}1}\sum_{cycl}\frac{(a\minus{}b)^2}{(a\plus{}c)(b\plus{}c)}$
We have
$ \sum_{cycl}\frac{bc\minus{}a^2}{a^2\plus{}bc}\equal{}\frac{1}{2}\sum_{cycl}\frac{(a\plus{}c)(b\minus{}a)\plus{}(a\plus{}b)(c\minus{}a)}{a^2\plus{}bc}
\equal{}\frac{1}{2}\sum_{cycl}\frac{(a\minus{}b)^2(ab\minus{}c^2)}{(a^2\plus{}bc)(b^2\plus{}ca)}$
Thus the original inequality is equivalent to
$ \sum_{cycl} (a\minus{}b)^2\left(\frac{2}{(a\plus{}c)(b\plus{}c)}\plus{}\frac{(2k\plus{}1)(c^2\minus{}ab)}{(a^2\plus{}bc)(b^2\plus{}ca)} \right) \ge 0$
For $ 0\le k\le\frac{1}{2}$, it suffices to show that
$ (a^2\plus{}bc)(b^2\plus{}ca)\ge ab(a\plus{}c)(b\plus{}c)$
which [i]extremely[/i] well-known  :)

Thank you very much, Arqady and Ductrung, for your nice solutions based on SOS method. :)

[quote="ductrung"]Prove that for all positive real numbers $ a,b$ and $ c$,
\[ \frac {bc}{a^2 \plus{} bc} \plus{} \frac {ca}{b^2 \plus{} ca} \plus{} \frac {ab}{c^2 \plus{} ab}\le \frac {1}{2k \plus{} 1}\left( 3k \plus{} \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)\]
where $ k \equal{} 1/2$.[/quote]
See here: http://canhang2007.wordpress.com/2009/11/16/inequality-71-p-h-duc/

[quote="urun"][quote="ductrung"]Prove that for all positive real numbers $ a,b$ and $ c$,
\[ \frac{bc}{a^2\plus{}bc}\plus{}\frac{ca}{b^2\plus{}ca}\plus{}\frac{ab}{c^2\plus{}ab}\le \frac{1}{2k\plus{}1}\left( 3k\plus{}\frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\right) \]
where $ k\equal{}1/2$.[/quote]

Hi [b]ductrung[/b],

Your statement is true also when $ k= \frac12+\sqrt{2}$. It is a beautiful problem ![/quote]

\[ \frac{bc}{a^2\plus{}bc}\plus{}\frac{ca}{b^2\plus{}ca}\plus{}\frac{ab}{c^2\plus{}ab}\le \frac{1}{2k\plus{}1}\left( 3k\plus{}\frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\right) \]
$ k= \frac12+\sqrt{2}$.

eq is:

$a=b,b=c(\sqrt{2}-1)$

[quote=ductrung]Prove that for all positive real numbers $ a,b$ and $ c$,
\[ \frac{bc}{a^2\plus{}bc}\plus{}\frac{ca}{b^2\plus{}ca}\plus{}\frac{ab}{c^2\plus{}ab}\le \frac{1}{2k\plus{}1}\left( 3k\plus{}\frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\right) \]
where $ k\equal{}1/2$.[/quote]

[img]http://s9.sinaimg.cn/middle/006ptkjAzy76rdlqvPy98&690[/img]


function
calculus
derivative
calculus computations

How would you explain how sharp a maximum of a function is? I've drawn the first derivative, but that's a very graphical way that can be made to look sharp by varying the scale. Could I use the second derivative or something similar?

Try finding the curvature at the maximum point.

2-D: $k=\left| \frac{f''(x)}{(\sqrt{1+f'(x)})^3} \right|$

3-D: $k=\frac{|\bar{r}' \otimes \bar{r}''|}{|\bar{r}'|^3}$

The definition of curvature, is $\frac{1}{R}$, where $R$ is the radius of the osculating circle at that point.

Thanks for your help.

I've never used curvature before, so do you know where I can read up on it?

[quote="tux"]I've never used curvature before, so do you know where I can read up on it?[/quote]
You can  try [url]http://mathworld.wolfram.com/Curvature.html[/url]. By the way, since at the point of maximum $f'(x)=0$, the curvature of the graph at the point of maximum coinsides with $|f''(x)|$, so Jimmy's suggestion is basically the same as your idea about using the second derivative.


Find every function $ f: \mathbb{R} \to \mathbb{R}$ satisfies: $ f(f(x) \plus{} y) \equal{} 2x \plus{} f( f(f(y)) \minus{} x ),\forall x\in\mathbb{R}$.

[hide]I believe that its solution is either very complicated or creative and weird![/hide]

[hide]Yeah, finally I found a solution for this one. It is neither complicated nor creative and weird  :P

Put $ y\equal{}\minus{}f(x)$ then $ f( f(f(\minus{}f(x))) \minus{} x ) \plus{} 2x \equal{}0$ so $ f(x)$ is surjective on $ \mathbb{R}$.

Now suppose that there exists $ y_1,y_2$ such that $ f(y_1) \equal{} f(y_2)$ then we have $ f(f(x) \plus{} y_1) \equal{} f(f(x) \plus{} y_2),\forall x\in\mathbb{R}$. However, $ f$ is surjective so $ f(x \plus{} T) \equal{} f(x),\forall x\in\mathbb{R}$ where $ T\equal{}y_1 \minus{} y_2$.

Come back to the assertion, the substitution $ x$ by $ x\plus{}T$ yields $ f(f(x) \plus{} y) \equal{} 2(x\plus{}T) \plus{} f(f(f(y)) \minus{} x)$, meanwhile $ T\equal{}0$, or $ y_1 \equal{} y_2$.

So $ f$ is injective, and so it is bijective.

Let $ x\equal{}0$ we have $ f(y\plus{}f(0)) \equal{} f(f(f(y)))$, but $ f$ is injective so $ f(f(y)) \equal{} y \plus{} f(0),\forall y\in\mathbb{R}$.

Let $ y\equal{}0$, we obtain $ f(f(x)) \equal{} 2x \plus{} f(f(0)\minus{}x)$ and hence $ x\plus{}f(0) \equal{} 2x\plus{}f(f(0)\minus{}x)$ or $ f(f(0)\minus{}x) \equal{} f(0) \minus{} x,\forall x\in\mathbb{R}$.

Thus $ f(x) \equal{} x,\forall x\in\mathbb{R}$.[/hide]

[quote]Find every function $ f: \mathbb{R} \to \mathbb{R}$ satisfies: $ f(f(x) \plus{} y) \equal{} 2x \plus{} f( f(f(y)) \minus{} x ),\forall x\in\mathbb{R}$.[/quote]
[b]Here is my solution in the test:[/b]

$ f(f(x) \plus{} y) \equal{} 2x \plus{} f(f(f(y)) \minus{} x)$, $ \forall x\in \mathbb {R}$ $ (*)$

Take $ y \equal{} \minus{} f(x)$ into $ (*)$, we have $ f(0) \equal{} 2x \plus{} f(f(f( \minus{} f(x))) \minus{} x)$, $ \forall x\in \mathbb {R}$, which implies that $ f$ is surjective on $ \mathbb {R}$. Thus, there is $ a$ such that $ f(a) \equal{} 0$. Take $ x \equal{} a$ into $ (*)$, we have:

$ f(y) \equal{} 2a \plus{} f(f(f(y)) \minus{} a)$, $ \forall y\in \mathbb {R}$. Since $ f$ is surjective, then we conclude:

$ x \minus{} 2a \equal{} f(f(x) \minus{} a)$ $ (1)$, $ \forall x\in \mathbb {R}$. Now suppose there are $ x_1,x_2$ such that $ f(x_1) \equal{} f(x_2)$. From $ (1)$, it follows immediately that $ x_1 \equal{} x_2$.
Therefore, $ f$ is injective on $ \mathbb {R}$ (Then $ a$ is unique.)

On the other hand, take $ y \equal{} \minus{} a$ into $ (*)$, we also have:

$ f(f(x) \minus{} a) \equal{} 2x \plus{} f(f(f( \minus{} a)) \minus{} x)$, $ \forall x\in \mathbb {R}$ $ (2)$. Hence, combine $ (1)$ and $ (2)$, we conclude:

$ \minus{} x \minus{} 2a \equal{} f(f(f( \minus{} a)) \minus{} x)$, $ \forall x\in \mathbb {R}$ or $ f(x) \equal{} x \minus{} 2a \minus{} f(f( \minus{} a))$, $ \forall x\in \mathbb {R}$ $ (**)$.

But take $ x \equal{} y \equal{} 0$ into $ (*)$, we have $ f(f(0)) \equal{} f(f(f(0)))$, since $ f$ is injective, we have $ f(0) \equal{} f(f(0))$, use $ f$ injective, we conclude: $ f(0) \equal{} 0$. Therefore, $ a \equal{} 0$. From that combine with $ (**)$, we have: $ \boxed{f(x) \equal{} x}$, $ \forall x\in \mathbb {R}$.

[hide]Mashimaru's solution is also solutions of my two friends (Nhan and Thao)[/hide]

$ f(f(x) \plus{} y) \equal{} 2x \plus{} f(f(f(y)) \minus{} x)$, $ \forall x\in \mathbb {R}$ $ (*)$

Take $ y \equal{} \minus{} f(x)$ into $ (*)$, we have $ f(0) \equal{} 2x \plus{} f(f(f( \minus{} f(x))) \minus{} x)$, $ \forall x\in \mathbb {R}$, which implies that $ f$ is surjective on $ \mathbb {R}$. Thus, there is $ a$ such that $ f(a) \equal{} 0$. Take $ x \equal{} a$ into $ (*)$, we have:

$ f(y) \equal{} 2a \plus{} f(f(f(y)) \minus{} a)$, $ \forall y\in \mathbb {R}$. Since $ f$ is surjective, then we conclude:

$ x \minus{} 2a \equal{} f(f(x) \minus{} a)$ $ (1)$, $ \forall x\in \mathbb {R}$
put x=a we attain $ \minus{}a\equal{}f(\minus{}a)$ ;
put y=-a in  $ (*)$ we attain that
$ f(f(x) \minus{}a) \equal{} 2x \plus{} f(f(f(\minus{}a)) \minus{} x)$ we infer that x-2a=2x + f(-a-x) so f(x)=x-a $ \forall x\in \mathbb {R}$
and it is easy to find a=0
so f(x)=x $ \forall x\in \mathbb {R}$


My school is a public high school with a math and science academy inside of it.  For academy sophomores (like me), the school offers a field trip to florida every year, and it costs a lot of money, so payment starts very early.  I had already invested a lot of money in this camp, when I find out that my school will only offer the AMC 10/12 B this year.  Last year it only offered the A, so I was surprised, but more angry because the camp window falls during the AMC B date. (february 26-29 is the camp, 27th is the B date)  I can't back out of the camp because it's non-refundable, but I'm taking the AIME problem series B specifically for the AMC and AIME tests!  What do I do?

If you are willing to pay 60 dollars, you could just pay for overnight shipping for the AMC A since the deadline in Feb. 5.  There's no reason whatsoever that the school wouldn't let you pay for the whole package unless they're really stupid.

Or if you're doing USAMTS, you can hope to qualify for AIME through that.

As a last resort, you could try to find a university in Florida near your trip and ask them if you could take the AMC B there.


In the figure, polygons $ A$, $ E$, and $ F$ are isosceles right triangles; $ B$, $ C$, and $ D$ are squares with sides of length $ 1$; and $ G$ is an equilateral triangle. The figure can be folded along its edges to form a polyhedron having the polygons as faces. The volume of this polyhedron is

$ \textbf{(A)}\ 1/2\qquad \textbf{(B)}\ 2/3\qquad \textbf{(C)}\ 3/4\qquad \textbf{(D)}\ 5/6\qquad \textbf{(E)}\ 4/3$
[asy]
size(180);
defaultpen(linewidth(.7pt)+fontsize(10pt));

draw((-1,1)--(2,1));
draw((-1,0)--(1,0));
draw((-1,1)--(-1,0));
draw((0,-1)--(0,3));
draw((1,2)--(1,0));
draw((-1,1)--(1,1));
draw((0,2)--(1,2));
draw((0,3)--(1,2));
draw((0,-1)--(2,1));
draw((0,-1)--((0,-1) + sqrt(2)*dir(-15)));
draw(((0,-1) + sqrt(2)*dir(-15))--(1,0));

label("$\textbf{A}$",foot((0,2),(0,3),(1,2)),SW);
label("$\textbf{B}$",midpoint((0,1)--(1,2)));
label("$\textbf{C}$",midpoint((-1,0)--(0,1)));
label("$\textbf{D}$",midpoint((0,0)--(1,1)));
label("$\textbf{E}$",midpoint((1,0)--(2,1)),NW);
label("$\textbf{F}$",midpoint((0,-1)--(1,0)),NW);
label("$\textbf{G}$",midpoint((0,-1)--(1,0)),2SE);[/asy]

[hide="Click for solution"]
The polyhedron is a unit cube with a corner cut off.  The missing corner is a pyramid whose altitude is $ 1$ and whose base is an isosceles right triangle.  The area of the base is $ \frac{1}{2}$, so the pyramids volume is $ \frac{1}{3} \times \frac{1}{2}\equal{}\frac{1}{6}$, so the volume of the polyhedron is $ 1\minus{}\frac{1}{6}\equal{}\frac{5}{6}$, or $ \boxed{\textbf{(D)}}$.
[/hide]


Let $p$ be an odd prime.
Find all pairs of non-negative integers $x$ and $y$which satisfy
$p^x - y^p = 1$

[mod.: changed text size and title]

1) p=2, x=y=1

Edit 1:
Sorry i didnt read before that p must be an odd prime ...

Edit 2:
:)  Anyway p=3, x=2=y is also a solution...

Davron

[editing together of three posts by... well guess whom]

@Davron: please don't tripelpost in such small intervalls of time (just about three minutes for three posts here); edit the first post instead.
And please post proves instead of giving one answer that is easy to see.

there might be only one answer but i am unable to prove it.
i tried to see that $y$ is a multiple of $p-1$ but it doesn't work...

Rewrite ist as $p^x = y^p +1 = (y+1)(y^{p-1}-y^{p-2}+-...+1)$.
Thus $y+1=p^s$ with $s \geq 1$.
So we want to solve:
$p^x = (p^s-1)^p +1 = - \sum_{k=1}^{p}(-1)^k p^{ks} \binom{p}{k}$.
Now $p^{s+2}$ divides all summands except $(-p)^s \binom{p}{1}= \pm p^{s+1}$, so $s+1=x$ and also $\sum_{k=2}^{p} (-1)^k p^{ks} \binom{p}{k}=0$ is necessary. But the last one is impossible when $p \geq 5$ since then only the summand $p^{2s} \binom{p}{2} = \frac{p^{2s+1}(p-1)}{2}$ is not divisible by $p^{2s+2}$.
Thus $p=3$.
So we want to solve $3^x = (p \cdot p^{x-2}-1)^3+1 \geq 3^{3(x-2)}$, so $3x-6 \leq x \iff x \leq 3$.
Checking the remaining cases gives $x=2=y$ as the only solution.

this too appeared at a Czech olympiad... 1995, i think... looks like you're going through old olympiads and posting problems like mad, haha


A lot of people on AoPS have really odd usernames, that make no sense whatsoever (at least to me). Everyone use this thread to explain your username!

mine-read it in leet. LEETSPEAK4EVER

I take french, and my french name was mustafa in 8th grade. I like it a lot, so that is my username. This year, my high school french teacher told me that I had been spelling my name wrong the entire time, and that it should be mustapha, but I like mustafa much better.

Ragnarok= end of the world
23= my lucky number.

azn= for my asianness
demon= because i pwn everyone at my school
43= my score at chapter MC 2006, the time when I joined this forum.

funcia=my little club of eccentric friends

in one of my favorite books, a drow elf named Drizzt belongs to house Do'Urden.  House Do'Urden's ancestral name is [b]Daermon[/b] Na'Shezbarrnon (or something like that I forget)

the d stands for distance
so that gives distance runner. the 2007 stands for my graduation date.

[quote="junggi"][quote="i_like_pie"]i_like_pie = I like $\pi$ and pie.[/quote]
what about the "Principal of Inclusion Exclusion" aka PIE?[/quote]
Yeah, that too (that makes 3, my lucky number), but I didn't know about PIE till I joined AoPS. Which is why I would like to thank everybody who created AoPS, and all the people who constantly work on improving it. :yup:

name is neel. in an indian epic, nal was his twin brother.

well some usernames are pretty self explanatory (like "mike" and "1234567890") but some make no immediate sense like "Klebian" and "daine" and "bpms"

and mustafa....until his secret was revealed!

There have been [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=15608]several[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=79722]previous[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=107852]threads[/url] on this topic. You might find some of the names you're interested in there.

Oh, my username? Just my name; first initial plus half of last name.

I bet NOONE can figure out where my user name derives from!

key:it does [b]not[/b] have to do with red chickens eating hotdogs
 :rotfl:
-jorian

Hm....
jh = initial name
redsox = Boston Red Sox

:wink:

jorian=j  hm....
h=? skip it
jh could be initial name
redsox -> red sox/red socks... search it in wikipedia... exist
undoubtfully jh is initial name

:D

moogra - word I made up (definition was made up by this nasty person, but that was after I made all my accounts moogra)

I also use Froggy on other forums

There's kind of a story behind mine- I'm actually a trombonist.  However, once, my section was bored, as solofest had just finished and our director was like "we're not going to do anything productive today... is there anything you'd like to do?"  One of the other trombonists said "let's play baritones!"  So we did, and the power went out.  Then, a few weeks later, we looked in the band office, and found a trombone quartet called "T-Bone Party."  Instantly, one of us said "b-tone party" and the name's stuck ever since.

[quote="jhredsox"]oh, how could you of gotten it...i thought it was VERY discrete as i never use my name on posts :P or have ever said i like the redsox :P

... :rotfl:
-jorian[/quote]

http://www.onelook.com/?w=discrete
http://www.onelook.com/?w=discreet

A particularly annoying misuse to me.

I'm a Notre Dame football fan. Goldendomer=fighting irish supporters. :D

Guess where mine is derived from

[b]Hint:[/b] its from a website

PS-its amazing that this name wasn't taken yet[/b]

IT's from google! wow I'm smart, not really.

Wow, now a ranger. You are so smart :D

How can you know? :roll:

[quote="john lai"][quote="junggi"][quote="i_like_pie"]i_like_pie = I like $\pi$ and pie.[/quote]
what about the "Principal of Inclusion Exclusion" aka PIE?

junggi-my korean name. no idea WHY thats my username, honestly i hate it[/quote]

Want to know what junggi means in Korean (BTW do you know why koreans give you stange looks when you mention junggi)

PM me.[/quote]
I do know what it means(cmon im not a retard).
It's supposed to be "basis of government," so if you have something else in your mind, you're wrong.
"(BTW do you know why koreans give you stange looks when you mention junggi)"
wtf? looking at your username, you're not even Korean.
And you're probably pronouncing is very wrong.
and no one gives me strange looks bc of my name(explain how I survived through korean schools without being bullied severly).


MIT
college
Stanford
geometry
Princeton
Harvard
Science Olympiad

It is true that a lot IMO medalist get into top college, but that also because he/she also have other qualities(excellent grades, other participations on sport art...). So I am curious about what is purely the effect of one single IMO medal for a thirdworlc country(IMO week country) on his/her admition chance to top colleges. Consider the following:

Let P be a thirdworld student with following properties:
•	P is a IMO medalist or HM winner(let Q be the type of medal including HM)
•	P have no extracurricular other than math club, or only one art/sport activity in which P have no merit on it.
•	P have a class rank below 25%, but yet not bad GPA (due to third country education is easy so losing 0.1 point in GPA will let his class rank drop 15%)
•	P have no excellent standardized test scores, but yet good enough for most of common college

How likely is P get admitted into top colleges like MIT?
When Q varies from HM, Bronze, Silver and Gold, does the admitted chance considerably increase or just increase a little?

Well the EC's and grades will hurt, since those are the primary factors.  An IMO medal will help anyone a lot, but if they don't have the grades, scores, and  recommendations, nothing will them.

I don't work in admissions (thankfully :roll: ) but if I did, I would not be impressed by an applicant with a mediocre overall record and an IMO medal. It would make me think that the person has specialized in the IMO - and that is a very narrow specialty.

Without the other markers, the top colleges will dismiss the student immediately. At a lower-ranked college, the contest result may well be useless; they don't know what it means, since they haven't seen it before.

Actually, your scenario is unrealistic- the student will have excellent test scores. I can't imagine a student doing well on Olympiad-style contests and not blowing away something like the SAT. For short-answer contests, the high math/low verbal profile (English is a second language) is common, but Olympiads require writing skill.

To an admissions department, a low grades/high test scores profile is a gamble. That sort of student might pull things together and be inspired to do better in college, or might fail completely by not doing the work. A math contest background may be a good indicator here, since it indicates something academic the student really cares about.

mm... thanks to all...
but I didn't mean a guy with medal but sucked in everything else, but have no other specialities besides the medal

ok consider my own case:
 I currently study in Ecuador and have relatively good GPA 3.7 mightbe (here the grades are over 20 and I got 18.65 average) but yet low class rank top30%, I believe I could be at top 10% if just having 0.1 more in GPA, everyone's grade are too close together.
 I don't know my standardrized test grade but let assume a 90 in TOEFL, a 720 in sat II math and 700 in sat II physics and chemistry.
 no extracurriculars besides math club. (I don't think Wacraft III and Yu-Gi-Oh counts)
 I only have one IMO bronce medal.

 So, I want to know if I have good chance in getting into MIT. And I am more curious about if a IMO medalist successed in college application, is the IMO medal what played the biggest part? And if I could have a silver or gold medal, the chance to became admitted increase a lot or just a little?

Honestly, I don't see much of a chance.  MIT rejects top applicants, applicants with perfect SAT's, valedictorian, and pages of EC's.  Also, since you are international, if you are applying for financial aid, it will probably lower your chances more because usually, it is harder to get in because they have less money to give out to international students.

[quote="JRav"]MIT rejects top applicants, applicants with perfect SAT's, valedictorian, and pages of EC's.  .[/quote]

Out of interest, why do they do that? And what should you do if you have all of that (not that I have (or will apply there, or anything else related to that)) out of curiosity?

Awww......
so what I need to do know is study and get 3 big 800 in sat II(how difficult can it be)?
that will work?

Umm, absolutely not.  If anything, 3 800's shows them "Look, another overachiever who did nothing but study to try to impress us."  They would much rather see an applicant with a 2250 (still VERY good) ranked in the low 10% percentile who is also captain of 3 varsity sports and an Intel semifinalist (although that still might not be good enough!!!).

Someone in my school was 2 in the class, 2340 SAT, several SciOly medals.  Flat out rejected!!!

[quote="JRav"]
Someone in my school was 2 in the class, 2340 SAT, several SciOly medals.  Flat out rejected!!![/quote]

Sucks to be them... thats all I can really say.

Thus, the second piece of advice: Don't obsess over MIT. It's ridiculously hard to get into, and not significantly better for an undergraduate than a dozen other places that have less stringent admissions. You can get into plenty of good places- just likely not MIT.

Rather than worry about whether I would get into places, I just chose Berkeley (from in state)- and I don't think I would have had a better experience elsewhere.

[quote="jmerry"]Thus, the second piece of advice: [b]Don't obsess over MIT.[/b][/quote]

A friend of mine graduated from high school a year early and was admitted to MIT.  He tells me now that he thinks it would've been better just to go to a state school because

- it would be a lot cheaper, and
- AP scores could count for credit and he could skip most freshmen-level classes.

Another friend of mine at Stanford has been thinking the same thing, so I would like to reemphasize this point.  :)

[quote="t0rajir0u"]it would've been better just to [url=http://www.tamu.edu/]go to a state school[/url][/quote]  :)
There is an [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106786]excellent thread[/url] on the existence of life outside of Cambridge, MA. In any case, for an international applicant the issue of financial aid (or rather lack thereof) may well be a deciding factor.

Someone asked why MIT declines to admit certain amazing applicants, and the reply was

[quote]I suspect that is because the number of applicants they can admit is bounded above.[/quote]

I'm sure that's the main reason. Apply to more than one college if you are worried about this.

As someone who graduated from the "other Institute of Technology" with my degree in mathematics, I will give you advice that, at this present time, you will probably neither fully understand nor follow, but would be best served by doing both.

Do not believe that it is the institution that bestows upon you their grace and compliments if you are admitted; rather, it is always the other way around.  You do not serve them; your attendance and graduation is not something you owe anyone but yourself.  They serve you; it is they who are honored by your acceptance.

As such, you would be gravely mistaken if you believe that the name of the university you attend will confer upon you special privileges or particular prestige to the exclusion of all other facts.  You must choose the institution not upon the basis of its name or reputation, but upon how well it serves your needs--not just academically, but socially and financially as well.  I guarantee you that a 2.5 from MIT will not look anywhere near as nice as a 4.0 from UCLA, at least where graduate studies and future job prospects are concerned.  I also guarantee that you would be given more attention as an undergraduate mathematics major if you attend a smaller school with a more pedagogically motivated faculty, rather than a large research- and graduate-oriented institution, where the faculty care only about two things:  publication and tenure, leaving the actual teaching to TAs.  In this regard I am not singling out MIT or UCLA; it is a consideration that ought to be applied to every university.

Do not forget this.  Every professor who teaches you is there to serve YOU.  If you're unhappy with the quality of education or the amount of attention you are receiving, then you have made an uninformed decision, one that is all too common among students who apply to schools based on a perceived reputation or cachet which is primarily founded upon their research pedigree, and not on the actual quality of undergraduate education.

But I still fully expect you will continue to yearn for a spot at MIT.

Of course this is a question I really want to ask as well. First I want to thank all of you for the great answers provided.

This does not relate to the original question but just out of interest, say I study at a University in Australia until a certain degree (let's not say which yet). I want to do research in some area of Science. What happens when a person applies to do research at places like MIT, Princteton ro whatever?

Also some people I know really want to study oversea at places like UC Berkely, MIT, Cambrige, Oxford etc. Is the suggestion study one year at an Australian University and then apply for one oversea applicable at all (I don't think so but I want to confirm it).

I don't see any reason why one should study in their homeland first.  Universities will accept first-time international applicants, just make sure you're competitive.  As for research, I'm not sure what you mean by "apply for research".

[quote="BanishedTraitor"]say I study at a University in Australia until a certain degree (let's not say which yet). I want to do research in some area of Science. What happens when a person applies to do research at places like MIT, Princteton ro whatever?[/quote]

Undergraduates don't "apply to do research."  Some schools have undergraduate research programs (MIT and Caltech in particular), whereas at other schools the process is more informal (on a professor-by-professor basis).  Do you mean graduate study?  Graduate study is graduate study.

I really mean after University (maybe after obtaining a PhD or whatever).

Then you'd be applying to be a postdoc?  I don't really know much about what happens that high up the chain; that's years and years into the future, and you should probably think about what you're doing as an undergraduate first.

[quote="BanishedTraitor"]say I study at a University in Australia until a certain degree (let's not say which yet). I want to do research in some area of Science. What happens when a person applies to do research at places like MIT, Princteton ro whatever?[/quote]
They get rejected.  :maybe:

OK, not always. But the top postdoctoral positions are competitive, and those who did graduate studies in the U.S. have the advantage of being more visible. True, if you obtain a result that a major player at X takes a real interest in, then you have a chance of becoming a postdoc at X. :welcomeani: But your chances would be better if your started your career in the U.S. as a graduate student.

Based on talking to various admissions officers at top tier schools, etc, it seems to boil down to one issue: time.

I don't think that two 20-minute readings on an application is terribly thorough, but that's the minimum that many colleges promise. So let's do the math:
- That's equivalent to a 2/3 hour per application.
- Multiply that by, say, 12 thousand applicants, and you'll get 8000 hours of reading.
- Colleges will have about 10 weeks to go through these. So thats 800 hours of reading per week
- Or about 20 people working the average work week, just to see everything.
Two factors I haven't accounted for are the fact that some work longer than normal work weeks, and that it takes time to actually make the decisions, since admissions aren't based solely on the one person's application. All in all, that's cutting it real thin.

It's not that the valedictorian with the 2350+ and the Science Olympiad medals isn't good enough to make the limited class size. It's just that there's a limit to how accurately Admissions can pick the top people. In fact, they [i]don't need to[/i] pick the top people. They just need to pick a class with excellent students and diversity that will help them maintain their reputation.

mm...
interesting fact...
so in conclusion what all you guys said, IMO medallists don't get admitted unless he/she have every good stuffs like good grade leadership etc. which is enough to give him aceptance without the medal. this implies, the power of medal is minimal unless they need to decide between you vs. another guy with same profile but without medal... right?

As Zuton Force said, time is important factor, adm officer read too much application which he cannot remember all of them well, and eventually the lucky application that are able to print a deeper impression to officers are those who sucess??

[quote="JRav"]They would much rather see an applicant with a 2250 (still VERY good) ranked in the low 10% percentile who is also captain of 3 varsity sports and an Intel semifinalist (although that still might not be good enough!!!).
[/quote]

I don't think that MIT would consider the captain of 3 varsity sports to be a potential applicant since, at least in my impression, MIT places pretty low value on sports because it's not a major Division 1 or 2 school and doesn't have the best sports teams or emphasize athletics as much.  Also, in order to be captain of [b]3[/b] varsity sports would take tremendous energy, and unless your school sucks at sports, would be impossible to achieve while being an academically top student.  Consider five to six hours of practices every day and 9+ hours of sleep to recover.  Then 8 hours of school, 1 hour for meals, and you've got less than an hour to study.

Concerning the time and luck factor, it's always been well known that if your essay leaves a significant impression on the admissions officer by deviating from the normal essay themes and makes him/her think "Hmmm, after those 5000 admissions, this is the only one I remember," then you've pretty much increased your chances ten times.  But I'd think that an IMO medal would be pretty significant for an application, since it would make you one of only six in the entire country.

By the way we based most of these around MIT. Would the same apply for Princeton or UC Berkeley (and others)?

I'd think that if an IMO medal had an influence, it would work the best in an application for MIT, since MIT is much more well-known and better at math than other colleges.

[quote="ZzZzZzZzZzZz"]MIT is much more well-known and better at math than other colleges.[/quote]

Are you so sure?  Harvard is obviously more well-known than MIT, and I've heard it said that Princeton has the best math department in the country (possibly the world).  Do you mean that MIT is more sensitive to Olympiads?  I'm not so sure if that's true, either.

I'll speak to Berkeley: it's far easier to get into UC Berkeley, since it's a gigantic public school. My high school claims more than 30 students per year accepted there. It's probably somewhat harder if you're outside California, but there's a lot of room in general.
I don't know what a contest background would do- mine was irrelevant, since I was in automatically on grades and test scores.

A “top” college should know what an IMO medal means, even if it might not seek the medal as much as another top college.

I have a hard time believing that admissions officers from top colleges would give equal time to all applications. Many apps are likely subject to preliminary disqualification before the accompanying essays are half-heartedly scanned. That said, I cannot imagine an application with an IMO medal not getting extra, extra attention, even if the eventual outcome might be a rejection (due to unexplainable bad grades, inadequate recommendation, overall bad fit, etc.)

Assuming all top colleges in general demand a minimum standard on grades, test scores and recommendation -- will an admissions officer admit an IMO medal applicant who barely meets the standard over another applicant who is comfortably above the standard (tens of thousands of those each year) but otherwise shows little to suggest immense passion or potential for superior performance?

If an admissions officer from a certain college even ponders in the above scenario, then that college does not deserve an IMO medalist, period. No, an IMO medal is not just a tie-breaker. If it is then it can only serve that purpose 6 or less times a year :D.

With UC Berekely, however, I think 93% of the class is in-state California students, and often times, out-of-state students will not be admitted.

That's a really useless statistic in isolation.  Here's a more useful analysis: in fall of 2005, UC Berkeley had 37,014 applicants, of whom 9,829 were admitted and 4,105 attended. ([url=http://statfinder.ucop.edu/library/table_1c.aspx]cite[/url])  Of these, 30,850 (83%), 8,635 (88%) and 3,763 (92%), respectively, were California residents.  ([url=http://statfinder.ucop.edu/library/table_2c.aspx]cite[/url])  That means there was a 28% acceptance rate for California residents and a 19% acceptance rate for non-California residents, a fairly substantial gap, but that the makeup of the student body is also skewed towards Californians because the yield for California students is 44%, versus 28% for out-of-state students.  (This analysis could still be improved by taking into account some measure of the overall quality of the California versus out-of-state applicant pools.)

The numbers for the entire UC system are even more skewed against out-of-state students, but that's because nearly all out-of-state students apply to Berkeley or UCLA, the two campuses most difficult to get into.

[quote="JRav"]

Someone in my school was 2 in the class, 2340 SAT, several SciOly medals.  Flat out rejected!!![/quote]

A general statement on admissions:

Many times when discussing the chances of admittance for some academically high institution, students will give examples of other students with high GPAs and great SAT scores who got rejected. Usually when these stats are given they're used to discourage another student from applying to the school. This happens so much on websites like collegeconfidential that I no longer go to them. I like to think Aops is different.
If stats were the only factor in determining admission, stats would be the only piece of information colleges ask for. The practice of reducing an applicant to a few numbers that have been shown to have a merely tangetial correlation to aptitude is in my opinion absurd. I'm sure college admissions offices would agree. That's why they ask for recommendations and essays on how you've helped your community to see exactly what kind of student you are.


You are the ruler of a medieval empire and you will have a celebration tomorrow. You've got 1000 bottles of wine you were planning to open for the celebration, but you find out that one of them is poisoned. The poison shows no symptoms until death. Death occurs within ten to twenty hours after consuming even the tiniest amount of poison. You have thousands of prisoners who you can kill and just under 24 hours to determine which bottle is poisoned.
What is the smallest number of prisoners you must have to drink from the bottles so that you are absolutely sure to find the poisoned bottle within 24 hours?

So the poison kills a person within a random time from 10 to 24 hours later?

[quote="JBL"]Cute -- I think it's going to be something like [hide]$\log_{2}{1000}$[/hide][/quote]

That would work if there weren't a 24 hour time limit.

[quote="SplashD"]That would work if there weren't a 24 hour time limit.[/quote]

Why is that a problem?  We'll find out with 20 hours -- we have 24.  That gives me nearly 1/2 hour with each prisoner.

[quote="JBL"]Cute -- I think it's going to be something like [hide]$\log_{2}{1000}$[/hide][/quote]
That's right, except the answer is supposed an integer :P
good job, JBL!

So the answer is 10?

Can you explain how that works?

Ok. Assume the number of prisoners you kill is $k$ The strategy will be that each bottle is drunk by a distinct set of the $k$ prisoners, and if a certain set dies, we can then immediately find out the "bad" bottle. The number of sets that there can be formed from $k$ prisoners is obviously:
${k\choose 0}+{k\choose 1}+{k\choose 2}+...+{k\choose k}=2^{k}$
We have $2^{k}\ge 1000$, so $k\ge 10$
You can then easily assign each of $10$ prisoners a certain set of bottles to drink, and because $2^{10}\ge 1000$, this will be possible.

[quote="lightrhee"][quote="SplashD"]Say the king used 10 prisoners. How many different bottles would be used on each prisoners and which bottles?[/quote]
9 is impossible, since we have $2^{9}$ possible cases and that is not enough.

10 is possible and we can construct an example like this:
Assign a distinct number (1 to 1000) to all bottles. Convert all numbers into binary, and write all numbers with 10 digits(eg: 1 would be written as 000000001 and 510 will be 0111111110). The first prisoner drinks all bottles that has 1 as the first digit, the second prisoner drinks all bottles that has 1 as the second digit, and so on.
If prisoner #1,3,5,10 died, the poisoned bottle would have $1010100001_{(2)}$ as the assigned number.[/quote]

That's awesome. How did you think of coverting the bottle numbers into binary?

Let's add another $24$ bottles so that the total number of bottles is $1024$, let's assign numbers $1$ to $1024$ to the bottles.
Each prisoner will drink wine from $512$ different bottles
Let's number the prisoners as $1,2,...,10$
The king will have bottle $1$ drunk by nobody, each of bottles $2-11$ drunk by exactly one ditinct prisoner,..., each of bottles $\sum_{i=0}^{k-1}{10\choose i}+1$ to $\sum_{i=0}^{k}{10\choose i}$ drunk by a distinct set of $k-1$ prisoners.
Then if a certain set of prisoners dies, the bottle drunk by that set(we will have only one such bottle) will be the poisoned bottle.
(Removing the extra $24$ bottles and following the same strategy would result in the same thing)
EDIT: Awww, I type too slow
Nice solution, David!

[quote="SplashD"]That's awesome. How did you think of coverting the bottle numbers into binary?[/quote]

The basic notion is that each prisoner represents a single "bit" -- either he lives or dies.  That triggers the thought of binary.

This used to be an interesting "magic trick" (Very popular when I was a kid), you had a set of cards, (each card had few numbers on it)  and you asked the other person to  think of a number and select the cards  which has that number, you took a glance at the cards, and find-out the number.
Set of 10 cards covered all numbers from 1 to 1023..

[hide]Card 1 had -  1,3,5,7,9, .... (all numbers whose first bit is 1)
Card 2 had -  2,3, 6,7, 10,11, 14,15 .  (2 on 2 off IOW numbers whose  2nd bit is 1)
Card 3 had – 4,5,6,7  12,13,14,15, 20,21,22,23. (4 on  4 off )
Card 4 had – 8,9,10,11,12,13,14,15, 24,25, .. (8 on , then skip 8)
Etc…
Card 10 had - 512,513,514, ....
All you did  was to add up the first number of the cards selected.
(Eg if one had (card 1, card 3 and card 4) you added 1+4+8 = 13 is the number.[/hide]
Same  thing here,  all you have to see – which cards “died”.


I'd like some help getting a handle on this problem.  Is there a quick way to do it?

A triangle has sides 36, 40, and 52.  If the sides are increased by x, 2x, and 3x, where x is a positive integer, then the newly formed triangle is a right triangle.  Find the sum of all distinct possible perimeters of the newly formed triangle.

I'm assuming you meant the sides were increased respectively.  Otherwise the problem would become tedious.
The sides are 36+x, 40+2x, and 50+3x whcih yields $(36+x)^2=(50+3x)^2-(40+2x)^2$
$(50+3x)^2-(40+2x)^2=(10+x)(90+5x)=900+140x+5x^2$
$(36+x)^2=1296+72x+x^2$
Therefore $900+140x+5x^2=1296+72x+x^2$ Giving $4x^2+68x-396=x^2+17x-99=0$ which of course yields no integer x.
In this case, either your problem was incorrect or the increase of sides was not respective.  Either way you can solve the problem on your own using this method. :D

The increase was not intended to (necessarily) be respective.  I understand  the answer is 448; the problem appeared on a math contest exam intended to last 50 minutes.  Your solution method is the obvious approach, and the one I tried,  -- but I'm wondering if some shortcut exists?  I do know this much:  The side which ends up as the hypotenuse is not the same for all cases.

Pythagorean triples have the property that 3 divides at least one number, 4 divides at least one number, and 5 divides at least one number. Using 4 restricts x to even numbers. The rest is tedious casework, I guess. Just a bit better than brute force for all combinations.

I don't even know if this is worth saying, because it seems obvious, but it hasn't been said yet.  We just need to find the number of solutions and the sum of the solutions for x because the perimeter is 128+6x for each triangle.  128(#soln)+6(sum x)


Two sisters go up 40-step escalators. The older sister rides the up escalator, but can only take 10 steps up during the ride since it is quite crowded. The younger sister runs up the empty down escalator, arriving at the top at the same time as her sister. How many steps does the younger sister take?

I wonder if we can solve it by setting equation!I don't really get the official solution using the argument that I think is ambiguous. If anybody needs that solution, tell me and I'll post it to discuss. :wink:

Surely the only ambiguity involves whether the escalators travel at the same speed?

[hide="I think this is right"]
I think the time it takes for the older sister is the same as if she just went up a 30 step escalator standing still. Let the younger sister take n steps. Then it's as if she went up n-40 step escalator standing still. So n-40=30, n=70, is that right?[/hide]

[quote="dgreenb801"][hide="I think this is right"]
. Then it's as if she went up n-40 step escalator standing still. So s-40=30, s=70, is that right?[/hide][/quote]
Your answer is correct :) . But can you explain more detailed in this part? I don't get how you have: n-40? Is it the number of steps the younger woman was pushed down while she's running up because she moves in the opposite direction to the escalator?

Yes. If the younger girl went at a rate of 40 steps per the amount of time it took the sister to go up, she would end up making no progress, since the elevator would take her down at that same rate. Thus, she must outpace it by moving 30 steps per that amount of time more than the elevator to end at the same time. In other words, both of them are going up a 30 step escalator standing still, since that is the net rate (I know, it's sort of confusing).


calculus integration counting distinguishability

$x+y+z+a+b=12$ and $x,y,z,a,b$ are all positive, and note that $(6,2,1,1,2)$ is different from $(2,1,1,2,6)$, how many integral solutions are there to $(x,y,z,a,b)$?

[hide]Is it ${11\choose4}=330$? I did several numbers (a+b+c=5, a+b=6, etc.) and found a pattern that I knew before. I forgot how to explain it...

Maybe if you took the last digit b away to leave x+y+z+a, since what ever that equals will determine b (If x+y+z+a=9, b must be 3). There are 11 numbers to choose from since b>0 and is an integer, and four positions to choose from. That sounds like 11 choose 4, but I don't understand where the logic comes in at all. [/hide]

If I understand what n and r mean:

[hide]${16\choose4}=1820$. [/hide]

Okay now why does that work? Apparently it still has something to do with taking $b$ away.

Suppose you have a list of sixteen blanks: ----------------

You have a bag of twelve stars: ************

And four commas. ,,,,

Now, how many ways are there to put the twelve stars and four commas in the sixteen blanks?

For example, **,***,**,***,**

Note that the commas now partition the twelve stars into five groups. This is identical to the original problem.

This is a neat trick I learned at camp.

what if the commas are next to each other? wouldn't then one of the numbers be 0?  :?

[hide="Combinations with Unrestricted Repetitions"]Use combinations with unrestricted repetitions.

Its equal to:

$\binom{5+(12-5)-1}{4} = \binom{11}{4} = 330$[/hide]

[hide="The concept"]My teacher calls it "Cookies and Kids".  You have 5 distinguishable kids, in this case, $x$, $y$, $z$, $a$, $b$, and 12 non-distinguishable cookies to give to them.  Then you use the partitioning method to get $\binom{16}{4}$.

If each kid has to get a positive number of cookies, then that means each kid must get at least one cookie.  Okay, give each kid a cookie.  So now you have 7 cookies left.  Use the partitioning method to distribute those and get $\binom{11}{4}$.

Thinking of it this way may make more complicated problems of this type easier.  For example, suppose you want the number of nonnegative integer solutions to $a+b+x+y+z \le 12$.  Okay, in that case, say you have 5 kids and a sixth cookie bag for yourself.  You distribute 12 cookies to these 6 cookie bags, and then take your own cookie bag and take the cookies from it (which could be anywhere from 0 to 12).  Then you've just given fewer than 12 cookies to the kids.  Done: it's $\binom{17}{5}$.[/hide]

if a,b,c,x,y are all positive , it is same as finding

a+b+c+x+y=12-5=7

which is 11C4

[quote="dimak11214"][hide]1820[/hide][/quote]

from what i see, that's wrong...

Is it 11C4 or 16C4? Apparently, there's a big agreement on 11C4 (which is what I found at first, with a pattern).

I believe one answer is correct if only positive solutions are counted, and the other is correct if any non-negative solution is counted.


geometry
search
combinatorics unsolved
combinatorics

Given $ 3n$ points on the plane.The distance between any $ 2$ points no more than $ 1$.
Prove that one can divide this set of points in  $ n$ triplet sets such that The sum triangle's areas (triangles is forming by $ 3$ points from each set) less than $ \dfrac{1}{2}$

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1180306703&t=42377

Pierre.


Ok so i have this differential problem that i worked out but need to have my work checked, because i don't think i'm getting the right answer... here it is:

1)   If the side e of a square is increased by 1%, then the area is increased approximately by what?

this is what i had down for work...

[b]A = e^2, where a = e and a+de = e+.01
dA/de = 2e
dA = 2e*de
dA = 2e*(e+.01)
dA = 2e^2 + .02e[/b]

did i miss something?

sh0ebox


Calculate the pH of a buffer formed by mixing 85 mL of 0.13 M lactic acid with 95 mL of 0.15 M Sodium Lactate.


AMC AIME geometry ratio circumcircle USA(J)MO USAMO

[quote="Altheman"][size=167] Mock AIME 5 for 2007[/size]

[b]Rules[/b]

The rules are standard for the aime, and are written on the first page of the test. Just to reiterate,
1. The test is 3 hours long.
2. Only paper, pencils, pens, rulers, compasses, protractors, and graph paper are allowed during the test. No calculators are allowed.
3. The answers are 000 to 999 inclusive.
4. Answers such as 1 and 23 should be written as 001 and 023, respectively.

[b]Answer Submission[/b]

Answers should be PMed to me, Altheman. Although not required, I request that a short description of how you did each problem accompany your answers. I want this because it helps me give feedback (something that I like to do...) and it helps me see what type of solutions people are giving. Also I would like to know what answers are guesses and what answers are not. This changes my feedback significantly.

You should assume that the answers fit in the correct form. However, if you are sure that the answer form or the problem does not make sense, send your complaint with your answers.

[b]Deadline[/b]

I am not sure about the deadline yet...some time before the end of January would be nice, however, I may extend the deadline later.

[b]Comments[/b]

Difficulty: The first half of the test is similar or slightly easier than an average aime. However, I made the later parts of the test slightly more difficult than the actual AIME. My point of view is that one should take more difficult tests so the actual test will be easier.

Topic Distribution: I will not say exactly what the problem distribution is...you can see that when you take the test. I tried to make it as similar to the recent AIMEs as possible (in particular, the 2006-2007 tests).

Thanks: I am very greatful for the help I got from JSteinhardt. Not only did he come up with some clever problems, but he challenged me to revise my test another time. This helped me make the test more clear and, hopefully, will have avoided all careless errors.

Notation: All notation is standard. Everything should be pretty clear as I decided to include pictures for the geometry problems. Note that AB, etc denotes the length of the segment AB...no directed lengths or anything confusing like that.

 I spend a fair amount of time trying to make interesting problems. I hope that you will not only try to work on the problems during the intial 3 hours, but after you have submitted answers, you will work at some of the harder problems. I hope that you guys enjoy the test. Good luck and have fun!

The test is hidden here...I advise the use of the pdf because that has pictures. If anything is hard to see, just zoom in.

If you have any comments, just post.

[hide="Problems"]1. A triple elimination tournament takes place with 100 teams. A team is eliminated once they have lost 3 games and there are no ties. What is the maximum number of games that can take place during such a tournament?

2. If $x$ and $y$ are numbers such that $xy=1$ and $x+y=7$, then find $\frac{1}{x^{3}}+\frac{1}{y^{3}}$

3. Let $0<a<b<c<d<e<f<g$ be a geometric sequence of integers. Let $\star(x)$ denote the number of divisors of $x$. For example, $\star(6)=4$ because $1$, $2$, $3$, and $6$ are the divisors of $6$. If $\star(a)=7$, $\star(g)=13$, and $d-c=432$, then find $b$.

4. The happyfullness of a number is the ratio of the number of its even divisors to the number of its odd divisors. For example, the happyfullness of $6$ is $1$ and the happyfullness of $7$ is zero. If I pick a random integer between $1$ and $1000$, inclusive, then the expected happyfullness of the integer can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

5. Find the second largest prime factor of 1,599,999,999.

6. Let $A$, $B$, $C$, $D$, and $E$ be points in 3d space. Each pair of points is connected by a line segment (and the segments only intersect at their endpoints). A move consists of traveling to another point along a line segment. How many possible paths are there that: start at point $A$, move $5$ times, and end up at point $A$ again.

7. $\triangle ABC$ has $AB=3$, $BC=2007$, and $CA=2008$. Point $E$ is on $AC$ such that $\angle ABE=\angle EBC$. The parallel to $BC$ through $E$ hits $AB$ at $F$. If $BE$ hits $CF$ at $D$, then find $\frac{CD}{DF}$.

8. I roll a fair 10-sided die (with sides numbered one through ten) until I get a number greater than or equal to my previous roll. If the expected value for the number of rolls is $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers, find the remainder when $m+n$ is divided by 1000.

9. If $p(x)=x^{3}-5x^{2}+4x-1$ has roots $a$, $b$, and $c$, then find $\left(a^{2}-\frac{1}{a^{2}}\right)\left(b^{2}-\frac{1}{b^{2}}\right)\left(c^{2}-\frac{1}{c^{2}}\right)$.

10. Let $S=\{1003,1002,...,1\}$ be an ordered set. Define the operation: to 3-reduce, as follows: take 4 adjacent elements, $a_{k}$, $a_{k+1}$, $a_{k+2}$, and $a_{k+3}$, and replace those 4 elements with one element, $(a_{k}+a_{k+3}+\omega* a_{k+1}+\omega^{2}* a_{k+2})$, where $\omega=\frac{-1+i\sqrt{3}}{2}$. So $\{1,2,3,4,5\}$ could be 3-reduced into $\{5+2*\omega+3*\omega^{2},5\}$. If $S=\{k\}$ after being 3-reduced $334$ times, find the maximum value for the real part of $k$.

11. Let $\triangle ABC$ be a triangle with $AB=35$, $BC=50$, and $CA=45$. Two externally tangent circles, $w_{1}$ and $w_{2}$, (centers $O_{1}$ and $O_{2}$), are tangent at $D$ and pass through $B$ and $C$, respectively. $B$, $O_{1}$, $O_{2}$, and $C$ lie on a line in that order. Let $w_{1}$ and $w_{2}$ intersect $AB$ and $AC$ again at $F$ and $E$ respectively. If $B$, $C$, $E$, and $F$ lie on a circle, find $AD^{2}$.

12. Let $x,y,z$ be distinct real numbers such that
\[x^{3}-x(y^{2}+z^{2})=y^{3}-y(z^{2}+x^{2})=z^{3}-z(x^{2}+y^{2}) \]
If $(x,y,z)=(x_{0},y_{0},z_{0})$ is a solution such that
\[1+\frac{x_{0}+y_{0}+z_{0}}{(x_{0})(y_{0})(z_{0})}=x_{0}\]
then find the minimum value of $2(x_{0})^{2}+2(y_{0})^{2}+2(z_{0})^{2}$.

13. Let $w_{1}$ be a circle smaller than and internally tangent to $w_{2}$. A line $L$ is tangent to $w_{1}$ at $T'$ and hits $w_{2}$ at $A$ and $B$. If $AT'=7$ and $T'B=56$, find the maximum area of $\triangle ATB$.

14. $379$ is a prime. Let $\Omega(k)$ denote the smallest positive integer $n$ such that $379 | (k^{n}-1)$. A fair $378$-sided dice, with faces numbered $1,\ldots,378$, is rolled, and the result is $r$. Let $p$ denote the expected value of $\Omega(r)$. If $p=\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers, find the remainder when $m+n$ is divided by $1000$.

15. Let $w_{1}$ be a circle and $P$ be a point outside the circle. Points $A$ and $B$ are on $w_{1}$ such that $PA$ and $PB$ are tangent to the circle. A line through $P$ intersects $w_{1}$ at $C$ and $D$ such that $D$ is between $P$ and $C$. $AB$ hits $DC$ at $E$, the circumcircle of $ADP$ hits $PB$ again at $F$, and $AC$ hits $PB$ at $G$. Let $AE*EB=12$, $GB=10$, and $CD=DP$. If $PF=a\sqrt{b}-c$ where $a$, $b$, and $c$ are positive integers, and $b$ is not divisible by the square of a prime, then find $a+b+c$.[/hide][/quote]
About what difficulty is it?
Thanks a lot, I'm going to do it tonight!

[quote="Altheman"][b]Difficulty[/b]: The first half of the test is similar or slightly easier than an average aime. However, I made the later parts of the test slightly more difficult than the actual AIME. My point of view is that one should take more difficult tests so the actual test will be easier.
[/quote]

[quote="Iversonfan2007"]^^what, better than the real aime???![/quote]
 :maybe:

Wow this is a nice test.  At first I was like this is too easy, but then I started to do it.  I really like the problems, especially 2 with really nice solutions (to be discussed later).  Great job!

yeah i totally agree. I did ten and it took me a solid 2 hours. I'm surprised at the easiness of some and difficulty of several others. I like it. I don't like having to work too hard for the early questions.

I really like this Mock AIME , although it seems like the difficultly spread is rather large between adjacent problems (ie number 9 and 10).

Great job!

A few comments:

a) Thanks for the compliments...

b) I have gotten a few PMs about #3...The problem is correct, if you are getting a contradiction, then you are making a false assumption.

c) If you want to discuss solutions, just send me a PM.

d) I was particularly proud of problem #10, I thought it used a unique twist on an old idea....I suppose it is more difficult than some of the later problems, like #12, for example. Perhaps it should have been placed later.

e) Part of the reason for the difficulty curve is that there would be some problems to work on after the test. A test where you can solve every problem is not as useful as a learning tool as a test where you can solve some, but not all, of the problems.

I thought it was hard  :maybe:
Hopefully harder than the real AIME...
And I need to stop making stupid mistakes...

But some of the problems were really nice  :) nice job Alex

[quote="PenguinIntegral"][quote="mdk"]some really nice problems.[/quote]
Agree. Mildorf and Athleman's tests are much better than the real thing.[/quote]

Well Mildorf's tests are basically olympiad problems.

hmm...I have gotten about 8 submissions so far...the highest score being 10...i wonder if anybody will beat that?

Also the real AIME contains problems that are equivalent to easy usamo problems. Usually #14 and #15 have proof elements (remember the locus problem from last year's AIME II).

I very much recommend this Mock AIME, to anyone, really. If you can only get a few problems on the AIME, there are definitely ones you can do. If you can get 10+, there are definitely problems worth doing as well.

[quote="PenguinIntegral"]I have a few answers, but like always I'm too scared to turn them in...

When will solutions be posted? I have a few questions but I don't really want to work on the AIME until I'm sure I can get the solutions, otherwise I don't get any sleep for a few nights...[/quote]

as i have said already, i would like to discuss problems...i don't have any full solutions, but i may or may not outline all of them...

I'm going to do this RIGHT NOW, and if I haven't pmed you with answer by tuesday, somebody needs to yell at me and get me to stop being stupid and lazy.

if i have time, im gonna try this just to see what an AIME's like
im not sure if i have a chance of qualifying or not

Oh, man I forgot about this.  I'll send in the answers I have (which is probably like 1 correct one).

I'm sorry for being such a newb, but could you tell me how to pm someone?

There is a button on the bottom right hand side of a persons post that says PM.  If that doesn't work, go to the Private Messages tab at the top of the page, and click it, then click New Post and type your subject and message, as well as the username you want to send the PM to.

oh thanks a lot. wow its right there how could i have not seen it

[quote="MysticTerminator"]I believe #12 is incorrect[/quote]

Yea...I noticed that a while ago...Since I cannot edit my first post, very few people answered that question, and the fact that it is obvious what the change should be, I decided to leave it as is. Of course if it makes you feel better, the last line of question #12 should be reworded to be:

[hide]Find the maximum value of $k$ such that $2(x_{0})^{2}+2(y_{0})^{2}+2(z_{0})^{2}>k$ for all satisfactory triples $(x_{0},y_{0},z_{0})$[/hide]

[quote="MysticTerminator"]although that is not the fix I would have chosen

it would be so much tastier to just [hide="do the funky chicken"] replace "minimum" with "infimum"[/hide] woooo![/quote]

I would eat more chickens, but most people don't like chickens; especially chicken flavors such as bad, funky or projective. I prefer to eat "clearly stated" chicken.

[quote="solafidefarms"]So when in Feb. is this done?[/quote]

uhh...more towards the end...but if you want to take the test, I rather you take it sooner rather than later.

Can we still turn in answers?

*prints out the test*

yes

i have no real reason to not take answers...

p.s. highest score is a 14 so far...

those were some nice problems, but I had no clue on the last one

I'm planning on doing it tomorrow, because I don't have school, so don't make the cutoff before then (not that that's very likely at this point)...

altheman,
I gotta ask you--how do you come up with those problems? How long did it take?

Please I would like to know what the word ''mock '' means in the  expression '' Mock Aime '' or ''mock putnam '' etc.

     Thank you !

mathbel,
it means it's a practice aime made by someone other than the testmakers themselves

cooljoe that doesn't clear anything up at all - "A test not made by the testmakers" surely seems a little strange to you?. A Mock AIME is an AIME that was made by somebody on AoPS, not an official one that the AMC wrote. So this test has never and will never be given as a AIME to determine who makes USAMO, but the AIMEs [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45]here[/url] have been officially administered before.

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3751081_2007_mock_aime_5_by_alex_anderson]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)

PS.1. Answer key is [url=https://artofproblemsolving.com/community/c5h136609p772760]here[/url]
PS.2. [color=#f00]Wording to P12 has been corrected[/color].


A tank contains 40l of solution containing 2g of substance per litre. A liquid solution containing 3g of this substance per litre runs in at the rate of 4l /min and the well-stirred mixture runs out of the tank at the same rate. Find the amount of substance in the tank after 15minutes.

-----------------------
Let S(t) = amount of substance in tank at time t

then, $ \frac{dS}{dt}$=rate in - rate out

rate in $ \equal{}\equal{}\equal{}>$ $ \frac{3g}{L} . \frac{4L}{min}$

therefore, rate in = $ \frac{12g}{min}$

rate out ===> $ \frac{S(t)}{40min}.\frac{4L}{min}$

rate out = $ \frac{S(t)}{10min}$

so, $ \frac{dS}{dt}\equal{}12\minus{}\frac{S}{10}$

$ \frac{dS}{dt}\plus{}\frac{S}{10}\equal{}12$

Am I on the right track with this problem?


function
search
calculus
calculus computations

Does somebody have good problems involving periodicity of functions?

Why not using the resources of this very site?

If you go to Search and enter "periodic function", you get 415 matches. I agree, not all are relevant, good or interesting problems, but I'm confident you may cull a few...


inequalities
inequalities unsolved
algebra

It's austrian-polish 2000 and it seems easy
 Now, put $x=ab+bc+ca,y=abc$ then we get
 $(1-a^2)^2+(1-b^2)^2+(1-c^2)^2=2x^2+4y+2$ and
$(1+a)(1+b)(1+c)=2+x+y$ then ...
(notices: $x \leq \frac{1}{3}$ and $x^2 \geq 3y$ )

For L.H.S., it can be proved that if $f(a,b,c)=\sum (1-a^2)^2 - 2$, then $f(a,b,c) \ge f(\frac{a+b}{2},\frac{a+b}{2},c)$ Then it's easy to finish it.

[quote="s.m.a"]If three nonnegative reals $a$, $b$, $c$ satisfy $a+b+c=1$, prove that

$2 \leq \left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2}\leq \left(1+a\right)\left(1+b\right)\left(1+c\right)$.[/quote]
Put $A =x^{2}+y^{2}+z^{2}, B = xy+yz+zx, C = x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}, D = xyz.$ Then $1 = A+2B, B^{2}= C+2xyz(x+y+z) = C+2D$, and $x^{4}+y^{4}+z^{4}= A^{2}-2C = 4B^{2}-4B+1-2C = 2C-4B+8D+1.$

So middle expression $= 3-2A+(2C-4B+8D+1) = 2+2C+8D \ge 2$, with equality iff two of x,y,z are zero.

Right-hand expression$= 2+B+D.$ So we have to show that $2C+8D \leq B+D$ or$B-2B^{2}-3D \ge 0.$ By Cauchy-Schwartz, $A \ge B$, so $B(1-2B) = BA \ge B^{2},$ so it is sufficient to show that $B^{2}-3D = C-D \ge 0.$ But $C \ge xy\cdot yz+yz\cdot zx+zx\cdot xy = D$ by Cauchy-Schwartz.

[quote="s.m.a"]If three nonnegative reals $a$, $b$, $c$ satisfy $a+b+c=1$, prove that

$2 \leq \left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2}\leq \left(1+a\right)\left(1+b\right)\left(1+c\right)$.[/quote]

See also http://www.kalva.demon.co.uk/aus-pol/apsol/asol009.html , and there is a weaker version of the left part at http://www.mathlinks.ro/Forum/viewtopic.php?t=27430 . For what it is worth, let me present a different solution:

[color=red][b]Left part:[/b][/color] We have to prove that, if a, b, c are nonnegative numbers satisfying a + b + c = 1, then $2\leq\left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2}$.

Indeed, note that $\left(1-a^{2}\right)^{2}=\left(\left(1+a\right)\left(1-a\right)\right)^{2}=\left(1+a\right)^{2}\left(1-a\right)^{2}$. By the Bernoulli inequality, $\left(1+a\right)^{2}\geq 1+2a$; thus, since $\left(1-a\right)^{2}\geq 0$, we get $\left(1-a^{2}\right)^{2}\geq\left(1+2a\right)\left(1-a\right)^{2}$. Similarly, $\left(1-b^{2}\right)^{2}\geq\left(1+2b\right)\left(1-b\right)^{2}$ and $\left(1-c^{2}\right)^{2}\geq\left(1+2c\right)\left(1-c\right)^{2}$. But, as you can easily check, the identity

$\left(1+2a\right)\left(1-a\right)^{2}+\left(1+2b\right)\left(1-b\right)^{2}+\left(1+2c\right)\left(1-c\right)^{2}-\left(2+6abc\right)$
$=\left(a+b+c-1\right)\left(2a^{2}+2b^{2}+2c^{2}-2bc-2ca-2ab-a-b-c-1\right)$

holds, so that, since a + b + c = 1, we have a + b + c - 1 = 0 and

$\left(1+2a\right)\left(1-a\right)^{2}+\left(1+2b\right)\left(1-b\right)^{2}+\left(1+2c\right)\left(1-c\right)^{2}-\left(2+6abc\right)=0$,

and thus

$\left(1+2a\right)\left(1-a\right)^{2}+\left(1+2b\right)\left(1-b\right)^{2}+\left(1+2c\right)\left(1-c\right)^{2}=2+6abc\geq 2$.

Hence,

$\left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2}$
$\geq\left(1+2a\right)\left(1-a\right)^{2}+\left(1+2b\right)\left(1-b\right)^{2}+\left(1+2c\right)\left(1-c\right)^{2}\geq 2$,

qed.

[color=red][b]Right part:[/b][/color] We have to prove that, if a, b, c are nonnegative numbers satisfying a + b + c = 1, then $\left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2}\leq\left(1+a\right)\left(1+b\right)\left(1+c\right)$.

Using the cyclic sum notation, this inequality becomes $\sum\left(1-a^{2}\right)^{2}\leq\left(1+a\right)\left(1+b\right)\left(1+c\right)$.

Now, since a + b + c = 1, we have $\sum\left(1-a\right)=3-\left(a+b+c\right)=3-1=2$, and thus

$2\left(1+a\right)\left(1+b\right)\left(1+c\right)-2\sum\left(1-a^{2}\right)^{2}$
$=\sum\left(1-a\right)\left(1+a\right)\left(1+b\right)\left(1+c\right)-2\sum\left(1-a^{2}\right)^{2}$
$=\sum\left(\left(1-a\right)\left(1+a\right)\left(1+b\right)\left(1+c\right)-2\left(1-a^{2}\right)^{2}\right)$.

But

$\left(1-a\right)\left(1+a\right)\left(1+b\right)\left(1+c\right)-2\left(1-a^{2}\right)^{2}=\left(1-a^{2}\right)\left(1+b\right)\left(1+c\right)-2\left(1-a^{2}\right)^{2}$
$=\left(1-a^{2}\right)\left(\left(1+b\right)\left(1+c\right)-2\left(1-a^{2}\right)\right)=\left(1-a^{2}\right)\left(2a^{2}+bc+b+c-1\right)$,

and, using a + b + c = 1, we have

$2a^{2}+bc+b+c-1=2a^{2}+bc+\left(b+c\right)\left(a+b+c\right)-\left(a+b+c\right)^{2}=\left(a-b\right)\left(a-c\right)$,

so that

$\left(1-a\right)\left(1+a\right)\left(1+b\right)\left(1+c\right)-2\left(1-a^{2}\right)^{2}=\left(1-a^{2}\right)\left(a-b\right)\left(a-c\right)$,

and the above becomes

$2\left(1+a\right)\left(1+b\right)\left(1+c\right)-2\sum\left(1-a^{2}\right)=\sum\left(1-a^{2}\right)\left(a-b\right)\left(a-c\right)$.

Now, note that, since the nonnegative numbers a, b, c sum up to 1, they are all $\leq 1$, so their squares $a^{2}$, $b^{2}$, $c^{2}$ are $\leq 1$, too, and thus, the numbers $1-a^{2}$, $1-b^{2}$, $1-c^{2}$ are all nonnegative. Also, note that we can WLOG set $a\geq b\geq c$; then, $a^{2}\geq b^{2}\geq c^{2}$, so that $1-a^{2}\leq 1-b^{2}\leq 1-c^{2}$. Thus, we can apply the Vornicu-Schur inequality (Lemma 1 in http://www.mathlinks.ro/Forum/viewtopic.php?t=24555 post #9) with $x=1-a^{2}$, $y=1-b^{2}$, $z=1-c^{2}$, and get

$\sum\left(1-a^{2}\right)\left(a-b\right)\left(a-c\right)\geq 0$.

Because of the above, this implies

$2\left(1+a\right)\left(1+b\right)\left(1+c\right)-2\sum\left(1-a^{2}\right)\geq 0$,

so that $\sum\left(1-a^{2}\right)^{2}\leq\left(1+a\right)\left(1+b\right)\left(1+c\right)$. And the proof is complete.

This is probably not the simplest solution, but, at least, it nowhere requires calculating expressions of degree more than 3, so it's better than straightforward bunching ;) .

  Darij

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=602524]Generalization of 23rd Austrian-Polish MO 2000[/url]

[quote=s.m.a]If three nonnegative reals $a$, $b$, $c$ satisfy $a+b+c=1$, prove that

$2 \leq \left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2}\leq \left(1+a\right)\left(1+b\right)\left(1+c\right)$.[/quote]
[url]https://artofproblemsolving.com/community/c6h50689p321205[/url]

The Right Hand Side :

Let $ x=a+b , y=b+c , z=c+a $

The original proposition equivalent to
$ \sum x^2(y+z)^2 \leq (x+y)(y+z)(z+x) $

equivalent to $ \sum2x^2(y+z)^2 \leq 2 \Pi(x+y) $

equivalent to $ 2\sum x^2y^2 \leq \sum xy^3+\sum x^3y$

AM,GM is ok!!!


linear algebra
matrix
algebra proposed
algebra

Let $\delta_{i,j}=\left\{\begin{array}{lcl} 1 &,& i=j\\ 0 &,& i\ne j \end{array}\right.$ and  assume that    $\left( \begin{array}{cccc} a(1,1) & a(1,2) &\cdots & a(1,n)\\ a(2,1) & a(2,2) & \cdots & a(2,n)\\ \vdots &\vdots & \cdots & \vdots \\ a(n,1) & a(n,2) & \cdots & a(n,n)\\ \end{array}\right)$  are $n\times n$ real numbers  (  $n\times n$ real matrix) with  $a(1,1)\ne 1 .$ For $\nu\in \{1,2,...,n+1\}$ define the numbers $f_{\nu}(i,j)\;$  in the following way :  $f_{1}(i,j): =a(i,j) -\delta_{i,j}$ and
\[ \begin{array}{|c|} \hline \\ f_{\nu +1}(i,j)=f_{\nu}(i,j) -\displaystyle \frac{f_{\nu}(i,\nu)f_{\nu}(\nu,j)}{f_{\nu}(\nu,\nu)} \\ \\ \hline \end{array} \; , \; \; \; \nu \in \{1,2,...,n\}.  \]
[b] 1)  Find the table (matrix) [/b]    $\left(\begin{array}{cccc} f_{n+1}(1,1) & f_{n+1}(1,2) & \cdots & f_{n+1}(1,n)\\ f_{n+1}(2,1) & f_{n+1}(2,2) & \cdots & f_{n+1}(2,n)\\ \vdots & \vdots &\cdots & \vdots\\ f_{n+1}(n,1) & f_{n+1}(n,2) & \cdots & f_{n+1}(n,n)\\ \end{array} \right) \; .$

[b] 2) Prove that    [/b]   $\; \; a(i,j)=\delta_{i,j}+\sum\limits_{\nu=1}^{n}\displaystyle \frac{f_{\nu}(i,\nu)f_{\nu}(\nu,j)}{f_{\nu}(\nu,\nu)} \; \; ,\; \; i,j\in \{1,2,...,n\}, n\ge 2.$ [b] ( "Herbert Wilf identity") [/b]

[quote="Rust"]2) is equivalent that   $f_{n+1}(i,j)=0$.[/quote]
Indeed, right ! Thank you , but it must be proved .