$ \begin{tabular}{|c|c|}\hline\\Text&Text\\\hline\\Text&Text\\\hline\end{tabular}$

I want to build a table with boxed cells. Unfortunately, the simple construction with | and \hline doesn't work, as vertical bars after the second aren't full height. How can I get the proper fully-boxed look?

Also, it would be nice to have less white space above the contents of the cells, and more below.

The gap is caused by the \\ after \hline. Removing them gives
$ \begin{tabular}{|c|c|}
\hline
Text&Text\\
\hline
Text&Text\\
\hline
\end{tabular}$
though using, for example,  \renewcommand{\arraystretch}{1.5} gives a nicer result (see latex.png image below).

If you want the text top-aligned in a cell with space at the bottom then use p{[i]width[/i]} whch allows you to have vertical spacing in a cell
$ \begin{tabular}{|p{1cm}|c|}
\hline
Text\vspace{1ex}&Text\\
\hline
Text&Text\\
\hline
\end{tabular}$

Thanks. I was using p{width} anyway for the real thing; this was just a toy example of my problem.


function
real analysis
integration
advanced fields
advanced fields unsolved

Suppose $ h$ is an almost-everywhere continuous c.d.f. on a compact subset $ B \subseteq \mathbb{R}^n$, and $ X(\cdot)$ is a measure on the Lebesgue measurable sets of compact $ \mathcal{X} \subset \mathbb{R}^m$.  For each $ N \in \mathbb{N}$, let $ D_N$ be the set of integer points (lattice) over $ 2^N$ in $ B$, and Let $ h_N$ be the c.d.f. for a mass function over $ D_N$.  Let $ d: \mathcal{X} \times B \mapsto \mathbb{R}$ be strictly positive.  Show that if $ h_n \rightarrow h$ under the Levy-Prokhorov metric (http://en.wikipedia.org/wiki/Levy-Prokhorov_metric), then $ \int_{\mathcal{X}}{ [\int_B {d(x,\beta)d(h_n)(\beta)} \minus{} \int_B {d(x,\beta)d(h)(\beta)}]^2dX(x)} \rightarrow 0$

I got it.  I can post it, if anyone's interested.


vector
trigonometry
geometry unsolved
geometry

Let $AK, BL, CM$ be the altitudes of an acute triangle $ABC$.

And $9AK+4BL+7CM=0$ (the lengths are in vector form)

Prove that one of the angles in $ABC$ is $\frac{\pi}{3}$

Divide $4\vec{BL}$ and $7\vec{CM}$ into 2 component vectors: one parallel to $9\vec{AK}$, the other perpendicular to $9\vec{AK}$.
The vectors which is perpendicular to $9\vec{AK}$ must have sum $0$ because $9\vec{AK}+4\vec{BL}+7\vec{CM}=$ [u]$0$[/u].
So $4BL\sin C=7CM\sin B\Rightarrow 4(\frac{2S(ABC)}{b})c=7(\frac{2S(ABC)}{c})b\Rightarrow \frac{b}{2}=\frac{c}{\sqrt{7}}$
Similarly, we can get $\frac{a}{3}=\frac{b}{2}=\frac{c}{\sqrt{7}}$
$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}=\frac{3^{2}+2^{2}-\sqrt{7}^{2}}{2(3)(2)}=\frac{1}{2}=\cos \frac{\pi}{3}$


A rectangular garden 50 feet long and 10 feet wide is enclosed by a
fence. To make the garden larger, while using the same fence, its
shape is changed to a square. By how many square feet does this
enlarge the garden?

Thois current area is 500 square feet.
We have 120 feet oof fence, so making a square, we get 120/4=30 foot sides.
The area of the square is 900 square feet.
900-500= $ \boxed{400}$


Could someone proof? Thanks

$\frac{1+sinx+icosx}{1-sinx-isinx}=(tanx+secx)i ,x _E _R _and_ x_ different _ Pi/2+kPi$

$\frac{sin\alpha}{2}+\frac{sin2\alpha}{4}+\frac{sin3\alpha}{8}+...=\frac{2sin\alpha}{5-4cos\alpha}$

let's relatex that piece over there ;)

$\displaystyle \frac{1+sinx+icosx}{1-sinx-isinx}=(tanx+secx)i$ with $x \in \mathbb{R}$ and $x\not = \frac\pi2+2k\pi$ ($k\in \mathbb{Z}$)

$\displaystyle \frac{1+\sin x+i\cos x}{1-\sin x-i\sin x}=(\tan x+\sec x)i$ with $x \in \mathbb{R}$ and $x\not = \frac\pi2+2k\pi$ ($k\in \mathbb{Z}$)

In fact $\sin, \cos,\sec$ looks better than $sin, cos,sec$ :P

Rewriting we have:$\frac{1+\sin x+i\cos x}{1-\sin x-i\sin x}=(\tan x+\sec x)i=\frac{(\sin x+1)i}{\cos x}$

Or,
\begin{eqnarray*}
\cos x+\sin x\cos x +i\cos^2 x&=&(1- \sin x- i\sin x)(i\sin x+i)\\
\cos x+\sin x\cos x +i\cos^2 x&=&i\sin x+i-i\sin^2 x-i\sin x-i^2\sin^2 x-i^2\sin x\\
\cos x+\sin x\cos x +i\cos^2 x&=&i-i\sin^2 x+\sin^2 x+\sin x\\
\cos x+\sin x\cos x +i(\sin^2 x+cos^2 x)&=&i+\sin^2 x+\sin x\\
\cos x+\sin x\cos x +i&=&i+\sin^2 x+\sin x\\
\cos x(1+\sin x)&=&\sin x(1+\sin x)
\end{eqnarray*}

$(1+\sin x)\neq 0$ so the final ecuation is $\sin x=\cos x$ :)

But the question was Proof???  actually are we supposed to solve the equation :?:  :?:

No, my mistake. But after simplifying I didn't get a right answer. Am I missing something?  :?

Actually, taking $x=0$ the equality isn't true. It might be a writing mistake or something.


Support
USAMTS
MATHCOUNTS
percent
geometry
AMC
AMC 12

The continuing support of the USAMTS by our sponsors depends upon active participation by many students.  Please spread the word about the USAMTS by inviting your friends and classmates to participate!

wow, USAMTS is booming compared to 2005...

[quote="FantasyLover"][quote="gauss1181"][quote="AIME15"]I would ask people in my "advanced" geometry class, but they are incapable of scoring over 15 on the AMC 8, so...a decent score on USAMTS would be difficult...also they don't really like math at my school.  Like 90% of the people in MATHCOUNTS at my school (we have 70 kids, will be cut later) just go because their friends go, or parents forced.[/quote]
geez my school doesn't even do contests beside mathcounts and only like 1% of the kids are interested in math :o[/quote]

lol my school doesn't even do mathcounts
and im the only one whos interested in math  :rotfl:[/quote]

thats like my school last year....

now, since i transferred, we have mc!!! and a geometry class, wich is supposed to be 9th grade in my school district

[quote="Poincare"]at my school, everyone hates math[/quote]
Just like mine. USAMTS probably SHOULD help spread the math word.

Haha I don't think it works that way. My math teacher got the USAMTS flier in the mail and passed it around in our class. It got from one end of the room to the other end pretty quickly.  :lol:

yeah... put up posters or something...
my school don't like math... seriously...

Hmm...I'll go ask kids at the high school math team, and people at math circle.

I only know about....4 or 5 people out of 3700 students who I am sure is capable of mathematical proof.

Yes, and guess why? :)

[hide]
They haven't been exposed to it enough.
[/hide]

I constantly badger my friends, but they're all either too lazy or too busy.

As I live so far from America I do not know your competitions such as USAMTS but after reading the title I thought it's maybe an international competition. "spread the word"  :)  Anyway the forum is good, I keep reading it.

No, it's not just that.

Most just don't like math of those 3700.

Of those who do like math, they number at about...maybe 50 to 100.

Of those, most of them have been exposed to proofs.

However, most of them HATE proofs and actually maybe about 1/3 of them are purely in it for college applications; they don't care about where the formulas they use come from and why they work.

So only a handful.

[quote="katya"][quote="AIME15"]I would ask people in my "advanced" geometry class, but they are incapable of scoring over 15 on the AMC 8, so...a decent score on USAMTS would be difficult...also they don't really like math at my school.  Like 90% of the people in MATHCOUNTS at my school (we have 70 kids, will be cut later) just go because their friends go, or parents forced.[/quote]

wow, there's like 4 people who signed up for mathcounts in my school (and I'm the only one whose any good) :P[/quote]

That's okay, NOONE from my school signed up for MathCounts.

how many ppl do usamts anyway? and what percent get gold silver bronze HM etc.

On the USAMTS website, go and look up the statistics from past years.

I want to let my son go there next year. Please help me to choose the skill he will need to learn to be able to go there. Thank you.

I don't understand by what you mean by "going there".

The USAMTS is a mail-in competition. See [url=http://www.usamts.org]this[/url] for full information.

As to the skills your child would need, the USAMTS covers various subjects. Take a look at past problems on the site for more information.

Perhaps you posted in the wrong forum?

Is this competition international? As you can see, I'm in Singapore right now, so I was wondering if I could participate.

Unfortunately, due to the sponsorship of the competition, it is only restricted to US students.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=164&t=372013]here[/url].

However, no one is stopping you from doing the problems and gaining problem-solving insight, which is essentially the purpose of the competition.

That's sad. There are a lot of math lovers in my school, though they don't get the exposure, and many of the teachers here would appreciate it if we were able to participate. I mean, it does bolster the competitiveness of one, which most lack. (Except the over 75,000 population at Mathlinks.) But I think it's okay. I'll just solve the questions.  :wallbash:

Haha I don't think people at my school are smart enough for this including me, my highest score on amc12 is only 100.5

Hi ! I'm a new participer in this website I'm TAHA student at High School
I have good level in maths but I should try to be batter
I wish that we get nice time together

[quote="Heero Kirashami"]Most just don't like math of those 3700.

Of those who do like math, they number at about...maybe 50 to 100.

Of those, most of them have been exposed to proofs.

However, most of them HATE proofs and actually maybe about 1/3 of them are purely in it for college applications; they don't care about where the formulas they use come from and why they work.

So only a handful.[/quote]
Wow.
That's just... sad.

The problem is, many people are interested in math but does not do USAMTS because they think it's too hard

USAMTS is pretty hard, but it's not impossible if you spend a lot of time on it.

Yeah, it is hard but fun since I am a middle schooler

We have everything (AMC, MATHCOUNTS, etc) but it seems like not many people in our middle school seem to be interested...

Also, I don't think anyone in my school can even solve these problems...?


calculus
integration
trigonometry
function
calculus computations

Find th integral of dx / (9+x^2).

a) 3arctan(x/3) +c
b) 1/3arctan(x/3) + c
c) 1/9arctan(x/3) + c
d) 1/3arctan(x) + c
e) 1/9arctan(x) + c

A way of thinking:

$ \int\frac{dx}{1\plus{}x^2}\equal{}\arctan x\plus{}C.$

Based on that, $ \int\frac{dx}{a^2\plus{}x^2}$ should involve an arctangent.

Treat this as a physical problem. Assume that $ x$ and $ a$ are both measured with the same units of measurement - maybe they're both distances. General principle: a "transcendental function," by which I mean essentially anything other than a monomial, should take as its arguments pure unitless numbers. Hence, I don't want to see $ \arctan(x),$ as that would put a number with units inside the transcendental function arctangent. But $ \arctan\left(\frac{x}{a}\right)$ would be acceptable.

Guess that $ \arctan\left(\frac{x}{a}\right)$ is the antiderivative. Differentiate it to check. Discover that you need to divide by $ a$ to make the check work. Arrive at the formula

$ \int\frac{dx}{a^2\plus{}x^2}\equal{}\frac1a\arctan\left(\frac{x}{a}\right)\plus{}C$

Final check: If $ x$ and $ a$ are in meters, the the integral should be in $ \text{m}^{\minus{}1}.$ With that $ \frac1a$ there, it does have those units of measurement.

Of course it's far more efficient to have memorized the more general antidifferentiation rule - but this manner of thinking is one way to recover it, and also a way to help hold it in your memory once you have it.


calculus
integration
inequalities
function
Cauchy Inequality
real analysis
real analysis unsolved

Let $ f: [0,1]\to\mathbb{R}$ be a continuous function such that $ \displaystyle\int_0^1 f^{3}(x)dx \equal{} 0$.
 Show that
\[ \displaystyle\int_0^1f^{4}(x)dx\geq\frac {27}{4}\left(\int_0^1f(x)dx\right)^{4}.\]

Let's $ F_n \equal{} \int_0^1f^n(x)dx$, by Cauchy-Schwartz: $ F_1^2 \leq F_2$, with equality for $ f \equal{} k$ constant function. In this case using $ F_3 \equal{} 0$ we get $ k \equal{} 0$ and the conclusion is obvious true.
Suppose now that $ F_1^2 < F_2$, using again Cauchy inequality: $ (\int_0^1(a \plus{} f^2(x))f(x)dx)^2 \leq \int_0^1(a \plus{} f^2(x))^2f^2(x)dx, a \in R$ and from here after small computations:
$ \minus{}a^2(1 \minus{} \frac{F_1^2}{F_2}) \minus{} 2aF_2 \leq F_4$
We are looking for constant $ a$ that maximize the left member, treating this member as a function of second degree in variable $ a$. We get: $ a \equal{} \frac{2F_2}{\minus{}(1 \minus{} \frac{F_1^2}{F_2})}$ and the maximum value for left member: $ \frac{F_2^2}{1 \minus{} \frac{F_1^2}{F_2}} \equal{} \frac{F_2^3}{F_2 \minus{} F_1^2}$
So for this particular value of $ a$ we have the best inequality: $ F_4 \geq \frac{F_2^3}{F_2 \minus{} F_1^2}$
Now consider the right member as a function in $ F_2$, using derivates and find a minimum point at $ F_2 \equal{} \frac{3}{2}F_1^2$ for which the right member has value $ \frac{27}{4}F_1^4$
[I hope I made no stupid mistakes]
Equality holds for functions that satisfy: $ F_3 \equal{} 0$ and $ F_2 \equal{} \frac{3}{2}F_1^2$
However I'm wondering about an example of such function, which is non constant

Hello;

[quote="mareleG"]
$ \minus{}a^2(1 \minus{} \frac{F_1^2}{F_2}) \minus{} 2aF_2 \leq F_4$
[/quote]
I don't see how you get this inequality. When we expanding your integral inequality

[quote="mareleG"] $ (\int_0^1(a \plus{} f^2(x))f(x)dx)^2 \leq \int_0^1(a \plus{} f^2(x))^2f^2(x)dx, a \in R$ [/quote]

we will have $a^2F_1^2-a^2F_2-2aF_4\le F_6$. If $a\in \mathbb R^-$ , then  $a^2F_1^2-a^2F_2-2aF_2^2\le F_6$. But , your inequality is  $a^2F_1^2-a^2F_2-2aF_2^2\le F_4 F_2$.

[quote="mareleG"]and from here after small computations:
[/quote]

Can you put these computations please? Since I think you have used a wrong equality which is $F_{n+m}=F_nF_m$.

when i said Cauchy inequality, i meant it, but i missed the second integral sign $\int$, sorry about that :), is something like:

$ (\int_{0}^{1}(a+f^{2}(x))f(x)dx)^{2}\leq\int_{0}^{1}(a+f^{2}(x))^{2}dx \int_{0}^{1}f^{2}(x)dx, a\in R $

is this leading you to what i said ?
thx

For wich non constant function equality hold in this problem?


A U tube contains mercury ($\rho = 13.6 g/mL$), water, and oil ($\rho = 0.80 g/mL$). The height of water is $h_{2}= 34 cm$, and the distance between the two levels of mercury is $h_{3}= 6.5 cm$. Compute the new height $h_{3}'$ between the two levels of mercury when the oil is transferred from the left branch to the right branch. In which branch is the level of mercury lower?
(sorry for the low picture quality)

I did not think that much in this but it should be right
(bit confusing problem  :D , but its good, thanks)

[hide="my try"]First we have equal pressure criterion:

$h_{2}\rho_{W}g+h_{1}\rho_{O}g=h_{3}\rho_{M}g$

The oil corresponds to a mercury pillar of height $\Delta h$.
It is given by $\Delta h=\frac{p_{oil}}{g\rho_{M}}=\frac{h_{3}\rho_{M}g-h_{2}\rho_{W}g}{g\rho_{M}}=h_{3}-h_{2}\frac{\rho_{W}}{\rho_{M}}$

Now the displacement will be $\frac{\Delta h}{2}$ from each side.
When pouring the oil back in the other part the other side must raise
$\Delta h$ so totally there is a displacement of $\frac32 \Delta h$ for each side.
So $h'=|h_{3}-3\Delta h|$ if positive the right side is still higher.[/hide]

Some how I did not use $\rho_{O}$, strange

Your answer doesn't seem to be right. Do your calculations correctly.

Well, perhaps you could be more precise and not just saying, it is wrong.  :) .
Please try to comment my solution, and if you're not satisfied please post your own.
Do we know the density of water?

[hide=" my try"]You surely agree that we have this:
$h_{2}\rho_{W}g+h_{1}\rho_{O}g=h_{3}\rho_{M}g$

Now when we remove the oil there is a excess pressure which has to be leveled out.
The oil corresponded to a mercury pillar of height $\Delta h$. We also assume that the
change of volume on one side equals that on the other. (Incompressibility)
So when oil is removed the tight parts sinks $\frac12\Delta h$ while the left raise by the same amount. ([color=blue]agree?[/color])

Now when the oil is put back on the right side we have kind of the same situation. Again we have an excess pressure of $\Delta h$. This is leveled out in the same way: $\frac12\Delta h$ change on each side.  ([color=blue]agree?[/color])

So the total change is:
Right: Raise by $\Delta h$
Left: lowering by $\Delta h$

The question was: [i]Compute the new height h_{3}' between the two levels of mercury when the oil is transferred from the left branch to the right branch. In which branch is the level of mercury lower? [/i]

Since the difference $h_{3}$ of the mercury levels have changed by $2\Delta h$ the new difference
is given by $h'=|h3-2\Delta h|$, and the modulus sign is because we might expect the change to be greater than the initial difference. [i]In that case[/i] $h3-2\Delta h<0$ and we know also know the right side is lower.  ([color=blue]agree?[/color])

So what is $\Delta h$? First of all it should not necessarily depend on on $\rho_{O}$ or $h_{1}$ directly,but it depends on the product $\rho_{O}h_{1}$, and this is known by the first equation.

Let $p_{O}$ be the pressure of the oil on the oil-water surface.
A mercury height "equal" to that pressure is
$\Delta h=\frac{p_{O}}{g\rho_{M}}=\frac{h_{3}\rho_{M}g-h_{2}\rho_{W}g}{g\rho_{M}}=h_{3}-h_{2}\frac{\rho_{W}}{\rho_{M}}$

so $h'=|h_{3}-2\Delta h|=|2h_{2}\frac{\rho_{W}}{\rho_{M}}-h_{3}|=|-1.5\,\mbox{cm}|$
So the left side is now $h'=1.5\,\mbox{cm}$ higher.[/hide]

[quote="Color"]Well, perhaps you could be more precise and not just saying, it is wrong.[/quote]

Yes, you are right on that. Anyway, I have read your last post and your reasoning and answer are correct.


geometry
circumcircle
AMC
AIME
trigonometry
trig identities
Law of Cosines

Circle O is the circumcircle of triangle ABC. Length AC=8, BC=6. Point D is on the side AB such that AD=4, and BD=3. CD is extended to intersect the circle at point E. Find the perpendicular distnace from E to AB.

[color=cyan]This is really rather hard to be labelled a medium question.  It is a good problem, though.  Konstantin's, I believe.[/color]

This reminds me of a problem from AIME 2002.

But Joel, I don't think it's that hard....

Because CD is the bisector of angle C, E bisects arc AB, ergo the perpendicular from E to AB passes through the circumcenter O. Given the circumradius R, we can find ED via x*(2R-x) = 4*3 = 12, by the power of a point. It is trivial to find R from the law of cosines on any side and corresponding angle, followed by the extended law of sines - but I'm not going to do it!

[quote]Because CD is the bisector of angle C, E bisects arc AB[/quote]

Why does it follow that E bisects arc AB?

Angle C subtends arc AB, CD is the angle bisector; E is the extension to the circle. Angle ACE is half of angle C, so arc AE, which it subtends, is half of arc AB.

Actually, there are TONS of ways to do this problem:

1. Angle bisector theorem + Stuart + Similar triangles
2. Law of Cosines
3. Lots of Powers of a Point

Point is......

[hmm, i'll tell you what my point is after i think of it...]


For positive $ x$, $ y$, and $ z$ such that $ x\plus{}y\plus{}z \equal{} 1$ prove the following:

$ 15(x^3\plus{}y^3\plus{}z^3\plus{}xy\plus{}yz\plus{}zx)\plus{}9xyz \ge 7$

$ \Leftrightarrow 15(x^3\plus{}y^3\plus{}z^3)\plus{}15(xy\plus{}yz\plus{}zx)(x\plus{}y\plus{}z)\plus{}9xyz\geq 7(x\plus{}y\plus{}z)^3$
$ \Leftrightarrow 15\sum x^3\plus{}15\sum_{sym}x^2y\plus{}54xyz\geq 7\sum x^3\plus{}21\sum_{sym}x^2y\plus{}42xyz$
$ \Leftrightarrow 8\sum x^3\plus{}12xyz\geq 6\sum_{sym}x^2y$
It is OK because $ \sum x^3\plus{}3xyz\geq \sum_{sym}x^2y$ (Schur inequality) and $ 2\sum x^3\geq \sum_{sym}x^2y$


prove that  if   lax^2+ bx + c l is in [-1;1]  for all  x in [-1;1],  so  lcx^2 + bx+ a l is in [-1;1] for all x in [-1,1].

i write this problem  the last week but the solution isn't for this problem .
 thanks

[quote="greatestmaths"]prove that  if   lax^2+ bx + c l is in [-1;1]  for all  x in [-1;1],  so  lcx^2 + bx+ a l is in [-1;1] for all x in [-1,1]. [/quote]

That's manifestly false : take for ex. $ 2x^2\minus{}1$ .

In general :   $ \displaystyle \ \ \forall_{x\in [\minus{}1,1]}\  |\sum_{k\equal{}0}^n a_kx^k|\leq 1\ \Longrightarrow\  \forall_{x\in [\minus{}1,1]}\  |\sum_{k\equal{}0}^n a_{n\minus{}k}x^k|\leq 2^{n\minus{}1}$
 :cool:

can you prove this :
$ \displaystyle \ \ \forall_{x\in [\minus{}1,1]}\ |\sum_{k\equal{}0}^n a_kx^k|\leq 1\ \Longrightarrow\ \forall_{x\in [\minus{}1,1]}\ |\sum_{k\equal{}0}^n a_{n\minus{}k}x^k|\leq 2^{n\minus{}1}$
Cool

[quote="greatestmaths"]prove that  if  $  |ax^2+ bx + c | \in [-1;1] $ for all  $ x \in [-1;1],$  so  $  |cx^2 + bx+ a | \in [-2;2] $   for all  $ x\in [-1,1]$. [/quote]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=343687&hilit=iranian]Iranian MO 2009 2nd round - IMO Longlist 1982 P56[/url]


geometry
3D geometry
conics
ellipse
parabola
hyperbola

Is the intersection of a cone and a plane an ellipse even if the cone isn't right? If yes, how can I prove it?

See [url]http://en.wikipedia.org/wiki/Conic[/url]
It can also be a parabola or a hyperbola

Whether the cone is right doesn't matter. You can perform shearing to make it right.

Thanks, but I've seen this article before. What do you mean by shearing...?
I need to write a proof for the fact that if a plane intersects a cone, without intersecting its base or its apex, then the intersection is an ellipse. Can you give me a few clues?


Show that if 14 distinct integers are chosen from the sequence 100,101,102,103,..,123,124, there must be two of them whose difference is 4.

[hide]Let us assume that there exist no two numbers chosen with a difference of 4.  We know that for any integer chosen, there is at least one integer which differs from it by 4, and so for every integer picked, there must exist at least one not picked.  However, we have 14 picked and 11 not picked.  Therefore, we have a contradiction, and there must exist at least two with a difference of 4.[/hide]

I think the problem should have been 13 integers are chosen....


trigonometry
calculus
calculus computations

let $ n$ be natural number , prove that the equation $ tan(x)\equal{}x$ has a unique solution in $ (\minus{}\frac{\pi}{2}\plus{}n\pi,\frac{\pi}{2}\plus{}n\pi)$.
and $ x_n\equal{}\frac{\pi}{2}\plus{}n\pi\minus{}arctan(\frac{1}{x_n})$ to deduce that $ x_n\equal{}\frac{\pi}{2}\plus{}n\pi\minus{}\frac{1}{n\pi}\plus{}\frac{1}{n{\pi}^2}\plus{} o(\frac{1}{n^2})$.

[quote="tchebytchev"]let $ n$ be natural number , prove that the equation $ tan(x) \equal{} x$ has a unique solution in $ ( \minus{} \frac {\pi}{2} \plus{} n\pi,\frac {\pi}{2} \plus{} n\pi)$.
and $ x_n \equal{} \frac {\pi}{2} \plus{} n\pi \minus{} arctan(\frac {1}{x_n})$ to deduce that $ x_n \equal{} \frac {\pi}{2} \plus{} n\pi \minus{} \frac {1}{n\pi} \plus{} \frac {1}{n{\pi}^2} \plus{} o(\frac {1}{n^2})$.[/quote] Maybe you could start by defining $ f(x)\equal{} tan(x) \minus{}x$, and then for an interval us the Intermediate Value Theorem (IVF). (Have you come across that? I'm not sure what to post any more due to always posting the wrong bits of information!)

Once you have proved that a solution exists (i.e. change of sign) for uniqueness, assume two solutions exist which are different and look for a contradiction.

(What have you done so far?)

[quote="beans"][quote="tchebytchev"]let $ n$ be natural number , prove that the equation $ tan(x) \equal{} x$ has a unique solution in $ ( \minus{} \frac {\pi}{2} \plus{} n\pi,\frac {\pi}{2} \plus{} n\pi)$.
and $ x_n \equal{} \frac {\pi}{2} \plus{} n\pi \minus{} arctan(\frac {1}{x_n})$ to deduce that $ x_n \equal{} \frac {\pi}{2} \plus{} n\pi \minus{} \frac {1}{n\pi} \plus{} \frac {1}{n{\pi}^2} \plus{} o(\frac {1}{n^2})$.[/quote] Maybe you could start by defining $ f(x) \equal{} tan(x) \minus{} x$, and then for an interval us the Intermediate Value Theorem (IVF). (Have you come across that? I'm not sure what to post any more due to always posting the wrong bits of information!)

Once you have proved that a solution exists (i.e. change of sign) for uniqueness, assume two solutions exist which are different and look for a contradiction.

(What have you done so far?)[/quote]
yes it's true what you say ,but actually the problem is to find that $ x_n \equal{} \frac {\pi}{2} \plus{} n\pi \minus{} \frac {1}{n\pi} \plus{} \frac {1}{n{\pi}^2} \plus{} o(\frac {1}{n^2})$ after using the fact that : $ arctan(x)\plus{}arctan(\frac{1}{x})\equal{} \pi/2$ for $ x>0$ and $ tan(y)\equal{}y  \Longleftrightarrow arctan(y)\equal{}y$
for $ y$ in $ ( \minus{} \frac {\pi}{2} ,\frac {\pi}{2}),$ and  $ y\equal{}x\minus{}n\pi$

Try a series expansion around each of the poles; $ \tan(\frac{\pi}{2}\plus{}n\pi\plus{}t)\equal{}\minus{}\frac1{\tan t}\equal{}\frac{\minus{}1}{t\plus{}O(t^3)}\equal{}\minus{}\frac1t\plus{}O(t)$.
Depending on how much precision you want, you may need to find that coefficient of $ x$.


Let $a_j>0\ (j=1,2,\cdots,n,n+1)\ $.Prove the following inequality without using $A.M.\leqq G.M.\leqq H.M.$.

\[n\left(\frac{a_1+a_2+\cdots\cdots+a_n}{n}-\sqrt[n]{a_1a_2\cdots\cdots a_n}\right)\leqq (n+1)\left(\frac{a_1+a_2+\cdots\cdots+a_{n+1}}{n+1}-\sqrt[n+1]{a_1a_2\cdots\cdots a_{n+1}}\right)\]

Substitute $a_{n+1}=a^{n+1}$ and $a_1a_2\ldots a_n=b^{n(n+1)}$.
Then it becomes: $a^{n+1}+nb^{n+1}\geq (n+1)ab^n$ which is true because of identity:
$a^{n+1}+nb^{n+1}-(n+1)ab^n=(a-b)(a^n-b^n)+b(a-b)(a^{n-1}-b^{n-1})+\ldots +b^{n-1}(a-b)(a-b)\geq 0$
The last inequality is obviously true. I don't know whether you were looking for something like this.

Thank you for your reply,ondrob.

You are right.

Here is my solution.

Let $A_n=a_1+a_2+\cdots\cdots+a_n,B_n=a_1a_2\cdots\cdots a_n$,

we have $a_{n+1}=\frac{B_{n+1}^{n+1}}{B_n^{n}}$,

so we can rewrite our inequality to $A_n- nB_n \leqq A_n+\frac{B_{n+1}^{n+1}}{B_n^{n}}-(n+1)B_n\Longleftrightarrow\left(\frac{B_{n+1}}{B_n}\right)^{n+1}\geqq(n+1)\frac{B_{n+1}}{B_n}-1$,

which is to be proved by Bernoulli's inequality,we are done.

This is exactly the same as rado's inequality, proved here: http://www.mathlinks.ro/Forum/viewtopic.php?t=21710


In the picture, what's the value of x ?

[img]http://img513.imageshack.us/img513/3875/aschas.jpg[/img]

A) 6 degrees
B) 12 degrees
C) 18 degrees
D) 20 degrees
E) 24 degrees

Left:$ 180^\circ\minus{}(3x\plus{}4x)\equal{}180^\circ\minus{}7x$,
Right:$ 180^\circ\minus{}(2x\plus{}6x)\equal{}180^\circ\minus{}8x$.
So, $ 5x\plus{}(180^\circ\minus{}7x)\plus{}(180^\circ\minus{}8x)\equal{}180^\circ \iff 180^\circ\equal{}10x \iff x\equal{}18^\circ$.

Thus, $ \boxed{C}$.

Focus attention on　center triangle.

Top+Right:$ 3x\plus{}4x$,
Top+Left:$ 2x\plus{}6x$
So Top+Right+Left:$ (3x\plus{}4x)\plus{}(2x\plus{}6x)\minus{}5x\equal{}10x\equal{}180^\circ \iff x\equal{}18^\circ$.

Thus, $ \boxed{C}$.

Very good :thumbup:
It's a good method, that i was difficult.
Thanks ;)


if y=x^3 then log y=3logx is this correct?

Yes that's right. In general,
$ log_a (b^c) = c  log_a b $

i think you should also add $x>0$ and $y>0$.

I donno, can there be a log with base equal to 0?

I don't think there is.
According to the definition of log function $y=\log_a{x}$, $a$ is a constant such that: $a>0,a\neq1$.


We have now made the various regional US communities open, in order to make it easier for users to join the discussion.  (We have not made the ones created for specific Math Circles or schools open - if you would like yours opened, please let me know.)

Thank you.  I was just wondering about this, because I wanted to use the AoPS At A Glimpse feature on the SC forum, but it was private.  Now I can use it!! Hooray!!


[hide]$ 12^2 \plus{} 16^2 \equal{} 400$
$ 24^2 \plus{} 10^2 \equal{} 676$
assuming $ a$ and $ b$ are integers:
$ a \equal{} 26, b \equal{} 20$ or vice versa, so:
$ a \minus{} b \equal{} 6$ or $ a \minus{} b \equal{} \minus{}6$ but I think there may be other solutions.[/hide]

is there an algebraic way without guess and check?

[quote="jimi"][hide]$ 12^2 \plus{} 16^2 \equal{} 400$
$ 24^2 \plus{} 10^2 \equal{} 676$
assuming $ a$ and $ b$ are integers:
$ a \equal{} 26, b \equal{} 20$ or vice versa, so:
$ a \minus{} b \equal{} 6$ or $ a \minus{} b \equal{} \minus{} 6$ but I think there may be other solutions.[/hide][/quote]

Jimi, did you get $ \ 12^2 \plus{} 16^2 \equal{} 400 , 24^2 \plus{} 10^2 \equal{} 676$ by guess and check or some other way?

Don't forget that it's possible that $ a\equal{}\minus{}26$, $ b\equal{}20$, and that makes it -46, so yeah, there are quite a few more solutions.

[quote="perfect628"]Well, I'm assuming he recognized the common Pythagorean triples $ (10, 24, 26)$ and $ (12, 16, 20)$. That would make sense.[/quote]

That makes sense.

Thank you. :lol:


trigonometry
algebra
polynomial
binomial theorem

Hey, I know this sounds pretty bad but i kinda messed up.This was my second attempt at IITJEE and AIEEE and i messed it up for some stupid reason.

Well, the thing is that when I was studying for IIT and AIEEE this year, I started to get drawn into math and i started liking it.I gave the CMI exam this year and answered 8 questions but i guess that wont be enough.

All I wanted to know is if theres anybody out there who know the number of attempts one has for the CMI and ISI exams and is there a limitation as to when thay can be given (eg IIT allows only 2 attempts one on the year of passing the boards and the other, the consecutive year.)

Thanks a lot for the advice

Hi, I have seen ppl appearing for ISI and CMI for the second time in my centre. But my guess is that they don't have any limit, maybe an age limit I guess. Better confirm it with your regional coordinator or the institutes itself.

Did you appear for ISI too?

Were u selcetd if u appeared?

[quote="realnachi"]What do they ask in the ISI interview?[/quote]

I dnt knw abt it

[quote="realnachi"]What do they ask in the ISI interview?[/quote]

well, few math profs. make you solve problems on a board

[quote="Scary math"]Are the interview questions like the ones u get in the written tests?[/quote]

yes. similar to them.

had my interview yestday .... got a problem that was a mix of trigo, complex nos and binomial theorem  :ninja:

Ummm.................
must be tough answering on spot! :( Is the difficulty level same as those of the written tests? :huh:
In the interview,is there any time limit sort of?I mean,do they give time to think,or u need to answer spontaneously? :huh:

What's your result?

[quote="Scary math"]Is the difficulty level same as those of the written tests? [/quote]
sort of, yes.

[quote="Scary math"]In the interview,is there any time limit sort of?I mean,do they give time to think,or u need to answer spontaneously? [/quote]
They give you lot of time to think n cross-question you till they're happy  :huh:
No time limit

[quote="Scary math"]What's your result?[/quote]
i cleared it

.

CONGRATS!!!!!!!!!!!!!!!!!!!!!! :thumbup:  :winner_first:  :first:

btw, the exact question was something like

can $ \sin nx$ and $ \cos nx$ be expressed as a polynomial in $ \cos x$ and $ \sin x$ respectively for all $ n \in N$ and to derive it  :)

should  $ \sin nx$ be only in terms of cosx . can;t the expression not have sinx ??

Sorry, typo there...

the Qsn should hav been -

can $ \sin nx$ and $ \cos nx$ be expressed as a polynomial in $ \sin x$ and $ \cos x$ ONLY respectively for all $ n \in N$ and to derive it.


After eating lunch together, Jay and Paul start walking in opposite directions. Jay walks 0.75 miles every 15 minutes and Paul power walks 2.5 miles every 30 minutes. In miles, how far apart are they after 1.5 hours?

change all speeds to mph

jay: .75*60/15=.75*4=3mph
paul: 2.5*60/30=2.5*2=5mph

after 1.5 hrs, their distance is

(5+3)*1.5=12

If we multiply the time and the distance on Jay's information by 6, we have Jay can walk 0.75*6=4.5 miles in 1.5 hoours.

If we multiply Paul's information by 3, Paul walks 7.5 miles every 1.5 hours. Therefore, after 1.5 hours they are 4.5+7.5=12 miles apart


When we have a problem

x,y,z>0. Prove that $ f(x,y,z)$ has smallest value when $ x\equal{}y\equal{}z$

Mixing variable))

when it's easy to show $ f(x,y,z) \ge f(\frac{x\plus{}y}{2},\frac{x\plus{}y}{2},z)$ for all x,y,z>0

Also $ f(x,y,z) \ge f(x,\frac{y\plus{}z}{2},\frac{y\plus{}z}{2})$ for all x,y,z>0

and $ f(x,y,z) \ge f(\frac{x\plus{}z}{2},y,\frac{x\plus{}z}{2})$ for all x,y,z>0

Then we can say $ f(x,y,z)$ has smallest value when $ x\equal{}y\equal{}z$ ???

Also if when it's not $ \frac{x\plus{}y}{2}$, but other mean such as $ \sqrt{xy}$

we can also say like that??

Hmm;; I want to use it when the problem is easy to prove such things f(x,y,z)>=f(x+y/2....),

and hard to prove f(a,a,b)>=M it.

I think it can be proven by check convergence of the sequence

$ X_{0} \equal{} x, Y_{0} \equal{} y, Z_{0} \equal{} z$

$ X_{n} \equal{} \frac {X_{n \minus{} 1} \plus{} Y_{n \minus{} 1}}{2}$

$ Y_{n} \equal{} \frac {X_{n} \plus{} Z_{n \minus{} 1}}{2}$

($ Z_{n} \equal{} Y_{n}$ for $ n \ge 2$)

[quote="arqady"][quote="dmath328"]When we have a problem

x,y,z>0. Prove that $ f(x,y,z)$ has smallest value when $ x \equal{} y \equal{} z$

Mixing variable))

when it's easy to show $ f(x,y,z) \ge f(\frac {x \plus{} y}{2},\frac {x \plus{} y}{2},z)$ for all x,y,z>0

Also $ f(x,y,z) \ge f(x,\frac {y \plus{} z}{2},\frac {y \plus{} z}{2})$ for all x,y,z>0

and $ f(x,y,z) \ge f(\frac {x \plus{} z}{2},y,\frac {x \plus{} z}{2})$ for all x,y,z>0

Then we can say $ f(x,y,z)$ has smallest value when $ x \equal{} y \equal{} z$ ???[/quote]
I don't think so. I think, there is a counterexample. But when you need it?

[quote="dmath328"]
Also if when it's not $ \frac {x \plus{} y}{2}$, but other mean such as $ \sqrt {xy}$

we can also say like that??[/quote]
The answer is the same.[/quote]
But I think, it is true  :maybe:

Of course, you need $ f$ to be continuous...

  darij


function
equation
IMO Shortlist
number theory
greatest common divisor

I would like to know if there is a formula for f(n^k) where n is a prime number.
I Think we can prove that f(ab)= f(a)f(b) when a and b are relatively prime.From here we can decompose every number in a product of powers of prime numbers and see how the funtion behaves.But i am really getting stucked here.  ;)

I showed that f(n^k) = (n-1)kn^(k-1) + n^k where n is prime and k naturel number (number of functions between two finite sets)
If we can prove a), then by writing every number as a product of powers of primes and using part a) we can reach the solution of part b).
For the complete proof ,many powers with indices are used and to type this am training on the Latex.

You can prove the multiplicativity easily by the chinese remainder theorem:

Let $x,y$ be coprime.

Then $\gcd(z,xy) = \gcd(z,x)\gcd(z,y) = \gcd(z \mod x,x)\gcd(z \mod y,y)$, and so $\gcd(r(a,b),xy))=\gcd(a,x)\gcd(b,y)$
(here $r(a,b)$ denotes the smallest natural number that fulfills $r(a,b) \equiv a \mod x,r(a,b) \equiv b \mod y$)

Since in
$f(xy)=\gcd(1,xy)+...+\gcd(xy,xy)$
and in
$f(x)f(y) = (\gcd(1,x)+...+\gcd(x,x))(\gcd(1,y)+...+\gcd(y,y)) =$
$= \gcd(r(1,1),xy) + \gcd(r(1,2),xy) + \gcd(r(1,3),xy) + ... + \gcd(r(1,y),xy) + \gcd(r(2,1),xy) + ... + \gcd(r(x,y),xy)$
every residue class $\mod xy$ (on the left side of the $\gcd$) occurs exactly one time, so they are identical.

Now use senouy's formula to get $f(2^{2n})=(n+1)\cdot 2^{2n}$, so a) is solved

Now $f(p^n)$ is only divisible by $p^n$ iff $p|n$ and then ($n=pk$) you get $\frac{f(p^n)}{p^n}=(p-1)k+1$
So if $n=sq$ with an odd prime factor $q$, set $r:=\frac{q-1}{2}$ and you will get $f(2^{2(s-1)}3^{3r})=n 2^{2(s-1)}3^{3r}$, so there are at least two solutions.

Let be $n=2^s$ now. For an odd prime $p$ also $f(p^k)$ is odd, so if $d$ is odd, by multiplicativity also $f(d)$ is odd.
Assume $f(2^id)=2^s \cdot 2^i d$, where $d$ is odd. Then by the formula for powers of primes and multiplicativity you get $2^{i-1}(i+2) f(d) = 2^s \cdot 2^i d \implies (i+2) f(d) = 2^{s+1} d$, but since $f(d)$ is odd, you need that $i+2 = 2^{s+1} m$ with some (odd) $m$. But this gives $m f(d) =d$, which is because of $f(x)>x$ for all $x>1$ only possible if $m=d=1$, so there is only one solution for this case. So also b) is solved.

As promised in Würzburg, I am posting my alternative solution to the problem.

[color=blue][b]Problem.[/b] Let $\mathbb{N}=\left\{1;\;2;\;3;\;...\right\}$ be the set of all positive integers. Define the function $f: \mathbb{N}\to\mathbb{N}$ as follows:

$f\left(n\right)=\sum_{i=1}^n\gcd\left(i;\;n\right)$.

[b](a)[/b] Prove that the function $f$ is multiplicative.
[b](b)[/b] Prove that for every positive integer $a$, the equation $f\left(x\right) = ax$ has a solution $x\in\mathbb{N}$.
[b](c)[/b] Prove that, for a positive integer $a$, the equation $f\left(x\right) = ax$ has exactly one solution $x\in\mathbb{N}$ if and only if $a$ is a power of $2$ (where $1=2^0$ is also considered as a power of $2$).[/color]

[i]Solution.[/i] First we note that $f\left(1\right)=\sum_{i=1}^1\gcd\left(i;\;1\right)=\gcd\left(1;\;1\right)=1$. For every $n > 1$, we have

$f\left(n\right)=\sum_{i=1}^n\gcd\left(i;\;n\right)>\gcd\left(n;\;n\right)=n$.

Now, in order to prove part [b](a)[/b], we recall some notations: For two functions $\alpha: \mathbb{N}\to\mathbb{N}$ and $\beta: \mathbb{N}\to\mathbb{N}$, the [i]Dirichlet convolution[/i] $\gamma=\alpha\bigstar\beta$ of $\alpha$ and $\beta$ is defined as the function $\gamma: \mathbb{N}\to\mathbb{N}$ such that

$\gamma\left(n\right)=\sum_{d\mid n}\alpha\left(d\right)\beta\left(\frac{n}{d}\right)$.

Further, we denote by $\nu: \mathbb{N}\to\mathbb{N}$ the function given by $\nu\left(n\right)=n$, and by $\phi: \mathbb{N}\to\mathbb{N}$ the Euler $\phi$ function.

Now consider the function $f$ defined by $f\left(n\right)=\sum_{i=1}^n\gcd\left(i;\;n\right)$.

Fix $n \in \mathbb{N}$. You can split the set of numbers $\left\{1;\;2;\;...;\;n\right\}$ into classes characterized by the greatest common divisor with $n$: For every number $k$ in the set $\left\{1;\;2;\;...;\;n\right\}$, the number $\gcd \left(k;\;n\right)$ is a divisor of $n$. Conversely, for every divisor $d$ of $n$, there are exactly $\phi\left(\frac{n}{d}\right)$ numbers $k$ in the set $\left\{1;\;2;\;...;\;n\right\}$ such that $\gcd \left(k;\;n\right) = d$ (in fact, there are exactly $\phi\left(\frac{n}{d}\right)$ numbers in the set $\left\{1;\;2;\;...;\;\frac{n}{d}\right\}$ which are coprime with $\frac{n}{d}$; multiplying them with $d$, we get exactly all numbers $k$ in the set $\left\{1;\;2;\;...;\;n\right\}$ such that $\gcd \left(k;\;n\right) = d$). Hence, the sum $f\left(n\right)=\sum_{i=1}^n\gcd\left(i;\;n\right)$ can be considered as a sum of divisors of $n$, with each divisor $d$ of $n$ occuring exactly $\phi\left(\frac{n}{d}\right)$ times. Hence, we can write

$f\left(n\right)=\sum_{d\mid n}\phi\left(\frac{n}{d}\right)\cdot d=\sum_{d\mid n} d\cdot \phi\left(\frac{n}{d}\right)=\sum_{d\mid n} \nu\left(d\right)\cdot \phi\left(\frac{n}{d}\right)$.

Thus, the function $f$ is the Dirichlet convolution of the functions $\nu$ and $\phi$. In other words, $f=\nu\bigstar\phi$. Since the functions $\nu$ and $\phi$ are multiplicative, it follows that the function $f$ is multiplicative, too. This solves part [b](a)[/b] of the problem.

Since the function $f$ is multiplicative, the function $g$ defined by $g\left(n\right)=\frac{f\left(n\right)}{n}$ must be multiplicative, too. In the above we found that $f\left(1\right) = 1$; thus, $g\left(1\right) = 1$. Also, in the above, we saw that $f\left(n\right) > n$ for every $n > 1$; this becomes $g\left(n\right) > 1$ for every $n > 1$.

For any prime p and any $a\geq 0$, we have

$f\left(p^a\right)=\sum_{d\mid p^a} d\cdot \phi\left(\frac{p^a}{d}\right)=\sum_{0\leq k\leq a} p^k\cdot \phi\left(\frac{p^a}{p^k}\right)$     (since all divisors of $p^a$ have the form $p^k$ for some k such that $0\leq k\leq a$)
$=\sum_{0\leq k\leq a} p^k\cdot \phi\left(p^{a-k}\right)=\sum_{0\leq k<a} p^k\cdot \phi\left(p^{a-k}\right)+p^a\cdot\phi\left(p^{a-a}\right)$
$=\sum_{0\leq k<a} p^k\cdot \phi\left(p^{a-k}\right)+p^a=\sum_{0\leq k<a} p^k\cdot p^{\left(a-k\right)-1}\left(p-1\right)+p^a$
$=\sum_{0\leq k<a} p^{a-1}\left(p-1\right)+p^a=ap^{a-1}\left(p-1\right)+p^a$,

so that

$g\left(p^a\right)=\frac{f\left(p^a\right)}{p^a}=\frac{ap^{a-1}\left(p-1\right)+p^a}{p^a}=\frac{a\left(p-1\right)+p}{p}$.

In particular,

$g\left(2^a\right)=\frac{a\left(2-1\right)+2}{2}=\frac{a+2}{2}$.

Hence, for every positive integer a, we have

$g\left(2^{2a-2}\right)=\frac{\left(2a-2\right)+2}{2}=a$.

Since $g\left(2^{2a-2}\right)=\frac{f\left(2^{2a-2}\right)}{2^{2a-2}}$, this yields $f\left(2^{2a-2}\right)=a\cdot 2^{2a-2}$. Thus, the equation f(ax) = ax has a solution $x\in\mathbb{N}$, namely $x=2^{2a-2}$ (we have $x\in\mathbb{N}$, since $2a-2\geq 0$, since $a\geq 1$). Hence, part [b](b)[/b] of the problem is solved.

Now remains to solve part [b](c)[/b] of the problem. Since $g\left(n\right)=\frac{f\left(n\right)}{n}$, the equation f(x) = ax rewrites as g(x) = a; hence, part [b](c)[/b] of the problem asks us to show that, for a positive integer a, the equation g(x) = a has exactly one solution $x\in\mathbb{N}$ if and only if a is a power of 2. In order to show this, it is enough to verify the following two assertions:

[i]Assertion 1.[/i] If a is a power of 2, then the equation g(x) = a has exactly one solution $x\in\mathbb{N}$.
[i]Assertion 2.[/i] If a is not a power of 2, then the equation g(x) = a has more than one solution $x\in\mathbb{N}$.

[i]Proof of Assertion 1.[/i] Let a be a power of 2. We have to show that the equation g(x) = a has exactly one solution $x\in\mathbb{N}$. This is trivial for a = 1 (in fact, g(1) = 1, while for any n > 1, we have g(n) > 1, so that x = 1 is the only solution of the equation g(x) = 1). Hence, in the following, it will be enough to consider the case when a > 1.

In order to show that the equation g(x) = a has exactly one solution $x\in\mathbb{N}$, we consider an arbitrary solution $x\in\mathbb{N}$ of this equation. If x is a power of 2, say $x=2^t$, then $a=g\left(x\right)=g\left(2^t\right)=\frac{t+2}{2}$, so that t = 2a - 2. Hence, we get the solution $x=2^{2a-2}$. Now, in order to show that this is the only solution of the equation g(x) = a, we have to show that if x is not a power of 2, then the equation g(x) = a cannot hold. And this can be showed as follows:

Since we have set that x is not a power of 2, we can write x in the form $x=2^t\cdot p_1^{a_1}\cdot p_2^{a_2}\cdot ...\cdot p_r^{a_r}$, where t is an integer $\geq 0$, where $p_1$, $p_2$, ..., $p_r$ are odd primes (i. e. primes > 2) and $a_1$, $a_2$, ..., $a_r$ are positive integers. Then, since the function g is multiplicative,

$g\left(2^t\cdot p_1^{a_1}\cdot p_2^{a_2}\cdot ...\cdot p_r^{a_r}\right)=g\left(2^t\right)\cdot g\left(p_1^{a_1}\right)\cdot g\left(p_2^{a_2}\right)\cdot ...\cdot g\left(p_r^{a_r}\right)$.

Since $x=2^t\cdot p_1^{a_1}\cdot p_2^{a_2}\cdot ...\cdot p_r^{a_r}$, we thus have

$g\left(x\right)=g\left(2^t\right)\cdot g\left(p_1^{a_1}\right)\cdot g\left(p_2^{a_2}\right)\cdot ...\cdot g\left(p_r^{a_r}\right)$.

Since $p_1^{a_1}>1$, we have $g\left(p_1^{a_1}\right)>1$. On the other hand,

$g\left(p_1^{a_1}\right)=\frac{a_1\left(p_1-1\right)+p_1}{p_1}$.

Since $p_1$ is odd, $p_1-1$ is even; thus, $a_1\left(p_1-1\right)$ is also even, so that $a_1\left(p_1-1\right)+p_1$ is odd again. Hence, the fraction $\frac{a_1\left(p_1-1\right)+p_1}{p_1}$ has an odd numerator and an odd denominator. In other words, the number $g\left(p_1^{a_1}\right)$ can be written as a fraction with odd numerator and odd denominator. We also know that this number is > 1. Similarly, same holds for the numbers $g\left(p_2^{a_2}\right)$, $g\left(p_3^{a_3}\right)$, ..., $g\left(p_r^{a_r}\right)$: They can also be written as fractions with odd numerators and odd denominators, and are > 1. Thus, the product $g\left(p_1^{a_1}\right)\cdot g\left(p_2^{a_2}\right)\cdot ...\cdot g\left(p_r^{a_r}\right)$ of all of these numbers is also a fraction with odd numerator and odd denominator, and is > 1. Multiplying this product with $g\left(2^t\right)=\frac{t+2}{2}$, we eventually kick some factors out of the denominator (in fact, some of the factors which divide t + 2), maybe we even get the number integer, but we surely won't get rid of the odd prime factors in the numerator. Hence, the number

$g\left(x\right)=g\left(2^t\right)\cdot g\left(p_1^{a_1}\right)\cdot g\left(p_2^{a_2}\right)\cdot ...\cdot g\left(p_r^{a_r}\right)$

is a fraction with odd prime factors in the numerator. Hence, it cannot be a power of 2 (greater than 1). Thus, since a is a power of 2 (greater than 1), the equation g(x) = a cannot hold. And thus, Assertion 1 is proven.

[i]Proof of Assertion 2.[/i] The proof of Assertion 2 will be mostly the same as given by ZetaX in post #4:

If a is not a power of 2, then a has an odd divisor > 1; in other words, we can write a in the form a = s (2r + 1), where $s\geq 1$ and $r\geq 1$ are integers (actually, $r\geq 1$ because the odd divisor 2r + 1 should be > 1). Then, because of the multiplicativity of the function g, we have

$g\left(2^{2\left(s-1\right)}\cdot 3^{3r}\right)=g\left(2^{2\left(s-1\right)}\right)\cdot g\left(3^{3r}\right)$
$=\frac{2\left(s-1\right)+2}{2}\cdot\frac{3r\left(3-1\right)+3}{3}=s\left(2r+1\right)=a$.

On the other hand, as we saw above,

$g\left(2^{2a-2}\right)=a$.

Since $2^{2\left(s-1\right)}\cdot 3^{3r}\neq 2^{2a-2}$ (in fact, since $r\geq 1$, the number $2^{2\left(s-1\right)}\cdot 3^{3r}$ is divisible by 3, while the number $2^{2a-2}$ is not), the equation g(x) = a thus has, at least, two solutions $x\in\mathbb{N}$. And Assertion 2 is proven.

This completes the solution of the problem.

[b]EDIT:[/b] Part [b](a)[/b] of this problem was also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=16599 .

  Darij

[quote="ZetaX"]You can prove the multiplicativity easily by the chinese remainder theorem:

Let $x,y$ be coprime.

Then $\gcd(z,xy) = \gcd(z,x)\gcd(z,y) = \gcd(z \mod x,x)\gcd(z \mod y,y)$, and so $\gcd(r(a,b),xy))=\gcd(a,x)\gcd(b,y)$
(here $r(a,b)$ denotes the smallest natural number that fulfills $r(a,b) \equiv a \mod x,r(a,b) \equiv b \mod y$)
[/quote]
 Can you show me how to get it or its proof

Then $\gcd(z,xy) = \gcd(z,x)\gcd(z,y)$ because $x,y$ are coprime.

$\gcd(z,x) = \gcd(z \mod x,x)$  and $\gcd(z,y) = \gcd(z \mod y,y)$ by Euklid's calculation of the $\gcd$, so by $\gcd(a,b) = \gcd(a-k\cdot b,b)$ for all $k$.

Then $\gcd(r(a,b),xy))=\gcd(a,x)\gcd(b,y)$ by setting $z=r(a,b)$ into the above.
(here $r(a,b)$ denotes the smallest natural number that fulfills $r(a,b) \equiv a \mod x,r(a,b) \equiv b \mod y$)

Is it clear now¿

Thanks for reply Zetax

[quote="ZetaX"]
$ f(xy) \equal{} \gcd(1,xy) \plus{} ... \plus{} \gcd(xy,xy)$
and in
$ f(x)f(y) \equal{} (\gcd(1,x) \plus{} ... \plus{} \gcd(x,x))(\gcd(1,y) \plus{} ... \plus{} \gcd(y,y)) \equal{}$
$ \equal{} \gcd(r(1,1),xy) \plus{} \gcd(r(1,2),xy) \plus{} \gcd(r(1,3),xy) \plus{} ... \plus{} \gcd(r(1,y),xy) \plus{} \gcd(r(2,1),xy) \plus{} ... \plus{} \gcd(r(x,y),xy)$
every residue class $ \mod xy$ (on the left side of the $ \gcd$) occurs exactly one time, so they are identical.
[/quote]

Sorry for asking again this is also mis understandable for me can you explain how did you show $ f(xy) \equal{} f(x)f(y)$
 Abdurashid

$ f(xy) \equal{} \gcd(1,xy) \plus{} ... \plus{} \gcd(xy,xy)$  by definition.

$ f(x) \equal{} (\gcd(1,x) \plus{} ... \plus{} \gcd(x,x))$ and $ f(y) \equal{} (\gcd(1,y) \plus{} ... \plus{} \gcd(y,y))$ also by definition, thus $ f(x)f(y) \equal{} (\gcd(1,x) \plus{} ... \plus{} \gcd(x,x))(\gcd(1,y) \plus{} ... \plus{} \gcd(y,y))$.

$ (\gcd(1,x) \plus{} ... \plus{} \gcd(x,x))(\gcd(1,y) \plus{} ... \plus{} \gcd(y,y)) \equal{}$
$ \equal{} \gcd(r(1,1),xy) \plus{} \gcd(r(1,2),xy) \plus{} \gcd(r(1,3),xy) \plus{} ...$
$ ... \plus{} \gcd(r(1,y),xy) \plus{} \gcd(r(2,1),xy) \plus{} ... \plus{} \gcd(r(x,y),xy)$
by the equality $ \gcd(r(a,b),xy)) \equal{} \gcd(a,x)\gcd(b,y)$ shown before.

Claim (then we are done):
$ \gcd(r(1,1),xy) \plus{} \gcd(r(1,2),xy) \plus{} \gcd(r(1,3),xy) \plus{} ...$
$ ... \plus{} \gcd(r(1,y),xy) \plus{} \gcd(r(2,1),xy) \plus{} ... \plus{} \gcd(r(x,y),xy) \equal{}$
$ \equal{} \gcd(1,xy) \plus{} ... \plus{} \gcd(xy,xy)$
To see this: when $ a \in \{ 1,2,...,x\}$ and $ b \in \{ 1,2,...,y\}$ vary, the value of $ r(a,b)$ gets everything from $ 1$ to $ xy$ exactly once;
this is true since: $ r(a,b) \equal{} r(c,d)$ iff $ a \equiv c \mod x$ and $ b \equiv d \mod y$, so (since $ a,c \in \{ 1,2,...,x\}$ and $ b,d \in \{ 1,2,...,y\}$) also iff $ a \equal{} c$ and $ b \equal{} d$.
This shows that every value is achieved once and only once, giving the desired equality.

I unnderstood your steps but the problem ıs hereç sorry for asking a lot
[quote="ZetaX"]$f(xy)=\gcd(1,xy)+...+\gcd(xy,xy)$  by definition.
$(\gcd(1,x)+...+\gcd(x,x))(\gcd(1,y)+...+\gcd(y,y))=$
$= \gcd(r(1,1),xy) + \gcd(r(1,2),xy) + \gcd(r(1,3),xy) + ...$
$... + \gcd(r(1,y),xy) + \gcd(r(2,1),xy) + ... + \gcd(r(x,y),xy)$
by the equality $\gcd(r(a,b),xy))=\gcd(a,x)\gcd(b,y)$ shown before.
[/quote]
here we have some terms two times as $\gcd(r(2,1),xy)$ is an example can you show how to substitute them , other purts ok.
Abdurashid

$g$ is mutiplicative function. a open question with $f(n)=\sum_{d|n}{g(gcd(m,n))}$

Note that $$\frac{f(n)}{n}=\sum_{d \mid n} \frac{\phi(d)}{d}$$ for all $n \in \mathbb{N}$. Thus, we see that $$\frac{f(n)}{n}=\prod_{p \mid n, \, p\,  \text{prime}} \left(1+v_p(n)\left(1-\frac{1}{p}\right)\right)$$ for all $n \ge 1$, solving (a). For each $a \ge 1$, put $x=4^{a-1}$, solving (b). For (c), we claim that any $n$ other than powers of $2$ admits multiple solutions to $f(x)=nx$.

Indeed, if $n \ne 2^k$ then there exists $p \mid n$ an odd prime and $q \in \mathbb{N}$ such that $n=pq$. Then $x=2^{2(q-1)}3^{3\left(\frac{p-1}{2}\right)}$ and $x=4^{n-1}$ are solutions of $f(x)=nx$.

If $n=2^k$ for $k \ge 1$, ($n=1$ is trivial), then for any $x \ne 4^{n-1}$ we have $$n=\frac{f(x)}{x}=\prod_{p \mid x, \, p\, \text{prime}} \left(1+v_p(n)\left(1-\frac{1}{p}\right)\right).$$ Note that $2 \mid x$ and $v_2(x)$ is even, else RHS is odd. Delete the factor $\left(1+\tfrac{1}{2}v_2(x)\right)$ as it is a power of $2$. Note the new RHS is not an empty product as $x \ne 4^{n-1}$. All the terms on the RHS are rational numbers with odd numerator and denominator, so we get a contradiction!

[quote=darij grinberg]
[b](a)[/b] Prove that the function $f$ is multiplicative.
[b](b)[/b] Prove that for every positive integer $a$, the equation $f\left(x\right) = ax$ has a solution $x\in\mathbb{N}$.
[b](c)[/b] Prove that, for a positive integer $a$, the equation $f\left(x\right) = ax$ has exactly one solution $x\in\mathbb{N}$ if and only if $a$ is a power of $2$ (where $1=2^0$ is also considered as a power of $2$).[/color]

[/quote]

Um... what about a=6?

[hide=Solution][quote=ISL 2004 N2]The function $f$ from the set $\mathbb{N}$ of positive integers into itself is defined by the equality \[f(n)=\sum_{k=1}^{n} \gcd(k,n),\qquad n\in \mathbb{N}.\]
a) Prove that $f(mn)=f(m)f(n)$ for every two relatively prime ${m,n\in\mathbb{N}}$.

b) Prove that for each $a\in\mathbb{N}$ the equation $f(x)=ax$ has a solution.

c) Find all ${a\in\mathbb{N}}$ such that the equation $f(x)=ax$ has a unique solution.[/quote]
[b][color=#000]Solution:[/color][/b] Since $m,n \in \mathbb{N}$ are such that $\gcd (m,n) =1$, thus,
\begin{align*} \gcd(k,mn) =\gcd( k,m) \cdot \gcd(k,n) \end{align*}
for all $k \le mn$. Recall that,
\begin{align*} \Psi : k \mapsto \left( k \pmod{m} , k \pmod{n} \right) \end{align*}
is a Bijection. Thus, using the fact that $\gcd(x,y) = \gcd(x-by,y)$ for all $b \in \mathbb{N}$ satusfying $x-by >0$, we get
\begin{align*} \left( \sum_{k=1}^m \gcd (k,m) \right) \left( \sum_{k=1}^n \gcd (k,n) \right) = \sum_{k=1}^{mn} \gcd (k,mn) \implies f(mn)=f(m) f(n) \end{align*}
This proves part (a). Experimenting with cases $a=1,2,3$ yields $x=1,4,16$. It's easy to establish
\begin{align*} f \left( 2^{k+1} \right) =2 f \left( 2^k \right) +2^k \implies f \left( 2^{2k} \right) =4 f \left( 2^{2k-2} \right) +2^{2k}  \end{align*}
from $f(1)=1$ $\implies$ $f \left(2^{2k} \right) =(k+1) 2^{2k} \implies$ there exists such $x \in \mathbb{N}$ for every $a$ $\in$ $\mathbb{N}$ proving part (b). Similarly, we may show for any prime $p$ and $n \in \mathbb{N}$,
\begin{align*} f \left( p^n \right) = (n+1) p^n -n p^{n-1} \end{align*}
Consider the case when $a$ has odd factors. We'll give construction showing that every such $a$ has atleast two solutions. Let \begin{align*} a = \prod p_i ^{f_i} \qquad, p_i <p_{i+1} \end{align*}
Choose $x$ with the same prime factors, thus
\begin{align*} x =\prod p_i ^{e_i} \implies a = \prod \frac{f \left( p_i ^{e_i} \right)}{p_i ^{e_i} } = \prod \left( e_i \left( 1 - \frac{1}{p_i} \right) +1 \right) \end{align*}
Now, the following clearly is a solution:
\begin{align*} \left( e_1,e_2,e_3, \ldots \right) \equiv \left( \frac{p_1 ^{f_1 +1} -p_1 }{p_1 -1}, \frac{p_2 ^{f_2 +1} -p_2 }{p_2 -1}, \frac{p_3 ^{f_3 +1} -p_3 }{p_3 -1}, \ldots \right) \end{align*}
From part (b), we always have a solution with only powers of $2$ for any $a$ which will be distinct than the above construction because for all $p_i \neq 2$
\begin{align*} e_i \left( 1 - \frac{1}{p_i} \right) +1 \qquad \text{ is odd} \end{align*}
Thus we get atleast $2$ solutions. When $a=2^k$
\begin{align*} 2^k = \left( \frac{e_1+2}{2} \right) \prod \left( e_i \left( \frac{p_i-1}{p_i} \right) +1 \right) \end{align*}
The lattter part is obviously odd and strictly greater than $1$, thus the $2$ in the denominator can only divide $e_1 +2$. Note
\begin{align*} 2^{ \nu _2 \left( \tfrac{e_1}{2} +1 \right) } = 2^k \end{align*}
Thus if we let $\tfrac{e_1}{2} +1$ $=$ $2^k \cdot d$. But, since $\prod \left( e_i \left( \frac{p_i-1}{p_i} \right) +1 \right)$ is strictly $>1$ which is impossible, thus$d=1$ $\implies$ $a=2^k$ works $\qquad \blacksquare$[/hide]

Solution from [i]Twitch Solves ISL[/i]:

Let $g(n) = f(n) / n$, so we are interested in the outputs of $g$. We start with:
[b][color=red]Claim:[/color][/b]  	The function $g$ is multiplicative 	and satisfies 	\[ g(p^e) = \frac{p-1}{p} \cdot e + 1 \] 	for any prime power $p^e$.
[i]Proof.[/i]  First, write \[ f(n) = \sum_{d \mid n} d\varphi(n/d) 	= \operatorname{id} \ast \varphi \] to get that $f$ is multiplicative (as the Dirichlet convolution of two multiplicative functions). Thus $g(n) = f(n)/n$ is multiplicative too.
Now note that for any prime power $p^e$, we have \[ g(p^e) = \frac{f(p^e)}{p^e} = 	\frac{p^e \cdot 1 + p^{e-1}(p-1) + \dots + 1 \cdot 	(p^e-p^{e-1})}{p^e} 	= e + 1 - \frac ep \] so the second part is true too. $\blacksquare$
In particular, we have \[ g(2^{2a-2}) = a \] so we already know every $a$ has the solution $x = 2^{2a-2}$.
We now show that this is the only solution if and only if $a$ is a power of $2$.
First, if $q > 1$ is any odd divisor of $a$, then writing $a = q \cdot b$, one can note that \begin{align*} 	g(2^{2b-2}) &= b \\ 	g(3^{\frac32(q-1)}) &= q \end{align*} and in this way we generate a new solution to the given equation. This shows the solution we found is not unique when $a$ is not a power of $2$.
Conversely, suppose $a = 2^\ell$ is a power of $2$ and $x$ is an integer with \[ g(x) = a = 2^\ell. \] Note that if $y$ is an odd prime power, then [list] 	[*]$\nu_2(g(y)) = 0$, and 	[*]$g(y) > 1$. [/list] So by measuring $\nu_2$, we get $\nu_2(g(x)) = \ell \implies \nu_2(x) = 2a-2$ matching the solution we found before. But then for size reasons, we must have $x = 2^{2a-2}$, as desired.

@v_Enhance loved the streams on Twitch they are awesome, I highly recommend them.  Thank you for the initiave.

By CRT, the multiset
\[\{\gcd(k,mn):1\le k\le mn\} = \{\gcd(a,m)\gcd(b,n):1\le a\le m, 1\le b\le n\},\]
so (a) follows.

Note that for a prime $p$,
\begin{align*}
f(p^e) &= (p^e-p^{e-1})\cdot 1+(p^{e-1}-p^{e-2})\cdot p+\cdots+(p-1)\cdot p^{e-1}+p^e \\
&= p^e\left[e(1-\tfrac{1}{p})+1\right],
\end{align*}
so we may now extend to all $n$ using (a). This also shows that $x=2^{2a-2}$ is a solution, proving (b).

We now need to find all $a$ for which there is no solution besides $x=2^{2a-2}$. Suppose $a$ has an odd factor $t\ge 3$. Then, $x=3^{3(t-1)/2}2^{2(a/t-1)}$ works, so we must have that no such $t$ exists, so $a$ must be a power of $2$.

In this case, suppose $f(x)=ax$ has a solution for $x$ where $p\mid x$ and $p\ge 3$. Let $e=v_p(x)$, and let $x=2^ty$ where $y$ is odd. We have a factor of
\[p^e\left[e(1-\tfrac{1}{p})+1\right],\]
in $f(y)$, which is odd and bigger than $1$, and since it divides $ax$, it must actually divide $y$. This shows that $f(y)\mid y$, which is a size contradiction unless $y=1$, so $x$ is also a power of $2$. It is easy to check that $x=2^{2a-2}$ is the only solution in this case. Thus the answer to (c) is all powers of $2$.

oops multiplicativity exists

$\textbf{Claim: }$ If $n=p_{1}^{e_1}p_{2}^{e_2}\dots p_{k}^{e_k},$ for primes $p_i,$ then $$\frac{f(n)}{n}=\prod_{i=1}^{k}\left(e_i+1-\frac{e_i}{p_i}\right).$$\\\\ $\emph{Proof: }$ Consider a prime $p\mid n$ with $\nu_p(n)=e.$ We have
\begin{align*}
    f(n)
    &= \sum_{d\mid n}(d\cdot\varphi(n/d))\\
    &= \sum_{d\mid n,\hspace{3pt} \nu_p(d)=e}(d\cdot\varphi(n/d))+\sum_{k=0}^{e-1}\left(\sum_{d\mid n,\hspace{3pt} \nu_p(d)=k}(d\cdot\varphi(n/d))\right)\\
    &= p^e\sum_{d\mid n/p^e}(d\cdot\varphi(n/d))+\sum_{k=0}^{e-1}\left((p-1)p^{e-1}\sum_{d\mid n/p^e}\right)\\
    &=p^{e-1}((e+1)p-e)f(n/p^e).
\end{align*} Repeating for all $p$ yields the desired statement. $\blacksquare$\\\\ a) Check that $x=2^{2a-2}$ works.\\\\ b) Since we want $\frac{f(x)}{x}$ to be an integer, we must have $e_i=m_{i}p_{i}$ for all $i.$ Then, $\frac{f(x)}{x}$ is a product of terms of the form $$m_{i}p_{i}+1-m_i=m_{i}(p_{i}-1)+1.$$ This expression is even if and only if $p_i=2.$ Thus, the solution to $f(x)=ax$ will be unique if and only if $a$ has no odd factors, i.e. $a$ is a power of $2.$

Let $Q(n)=\frac{f(n)}{n}$
Then $Q(n)=\frac{f(n)}{n}=\sum_{d|n} \frac{\phi(d)}{d}$ or $f(n)=n \sum_{d|n} \frac{\phi(d)}{d}$
a)Clearly $f(mn)=mn \sum_{d|n} \frac{\phi(d)}{d}=m \cdot n \cdot \sum_{d_1|n,d_2|m} \frac{\phi(d_1 d_2 )}{d_1 d_2}=n \sum_{d_1|n} \frac{\phi(d_1)}{d_1} m \sum_{d_2|m} \frac{\phi(d_2)}{d_2}=f(m) \cdot f(n)$
b)Since the function is multiplicative,it suffices to show the result for $X=p^e$
Plugging in gives,$Q(X)=\sum_{i=0}^e \frac{\phi(p^i)}{p^i}=\frac{p-1}{p} \cdot e + 1 $ and clearly this works for $N=2^{2(a-1)}$
c)The answer is all $a=2^F$ which clearly works.
If $a$ has an odd factor then $n=pq$. Then $x=2^{2(q-1)}3^{3\left(\frac{p-1}{2}\right)}$ and $x=4^{n-1}$ are solutions of $f(x)=nx$,a contradiction so we are done.

Observe that for some $d\mid n,$ we have $\gcd(k,n)=d$ if and only if $k=d\cdot l$ and $\gcd(l,n/d)=1.$ That is, for $\varphi(n/d)$ values. Thus \[f(n)=\sum_{k=1}^{n} \gcd(k,n)=\sum_{d\mid n}d\cdot\varphi\bigg(\frac{n}{d}\bigg)=(\text{id}*\varphi )(n).\]
For the first part, we will prove a stronger result which will, of course, solve our problem as well, since $\text{id}$ and $\varphi$ are multiplicative

[b][color=#3D85C6]Lemma:[/color][/b] For any multiplicative functions $f,g$ their Dirichlet Convolution, $f*g,$ is also multiplicative.

[b][color=#3D85C6]Proof:[/color][/b] Simply manipulate $f*g$ as follows: \[(f*g)(mn)=\sum_{d\mid mn}f(d)g\bigg(\frac{n}{d}\bigg)=\sum_{a\mid m}\sum_{b\mid n}f(a\cdot b)g\bigg(\frac{m}{a}\cdot\frac{n}{b}\bigg)=\]\[=\sum_{a\mid m}\sum_{b\mid n}f(a)f(b) g\bigg(\frac{m}{a}\bigg)g\bigg(\frac{n}{b}\bigg)=\sum_{a\mid m}f(a)g\bigg(\frac{m}{a}\bigg)\sum_{b\mid n}f(b)g\bigg(\frac{n}{b}\bigg)\]and the latter is clearly equal to $(f*g)(m)\cdot (f*g)(n).$ Hence, if $f$ and $g$ are multiplicative, then so is $f*g.$

Before moving on to the best parts, let's compute out function some more. Observe that \[f\big(p^k\big)=\sum_{i=0}^k p^i\cdot\varphi\big(p^{k-i}\big)=\cdots=p^k\bigg(\frac{p-1}{p}\cdot k+1\bigg).\]This already solves the second part of our problem, since $f(4^{n-1})=4^{n-1}\cdot n.$

Now, moving on to the third part, by the muliplicability of $f$ and the latter formula, we can deduce that \[\frac{f(n)}{n}=\prod_{p}\bigg(\frac{p-1}{p}\cdot\nu_p(n)+1\bigg).\]Assuming that $n=(2k+1)\cdot m$ for some $k>1$ then other than $f(4^{n-1})=4^{n-1}\cdot n$ we also have \[f(4^{m-1}\cdot 3^{3k})=\bigg(\frac{2-1}{2}\cdot (2m-2)+1\bigg)\bigg(\frac{3-1}{3}\cdot 3k+1\bigg)=n.\] Hence, if there exists a unique solution to $f(x)/x=n$ then $n=2^k.$ A simple check of $\nu_2$ will yield that indeed, all powers of two, have a unique solution.

a) By CRT, we know that for $k=1$ to $mn$, $k$ goes through all the different possible pairs of remainders mod $m$ and mod $n$ exactly once. Therefore, it is clear that $$f(mn)=\sum_{k=1}^{mn}\gcd(k,mn)=\sum_{k=1}^m\gcd(k,m)\sum_{k=1}^n\gcd(k,n)=f(m)f(n).$$
b) Let $g(n)=\frac{f(n)}{n}$. It is clear that $g(1)=1$ and that $g(mn)=g(m)g(n)$ for relatively prime $m,n\in\mathbb{N}$. We will look at powers of a prime $p$, say $p^k$. It is easy to show that $$f(p^k)=pf(p^{k-1})+\frac{p-1}{p}p^{k}$$ simply from how the definition of $f(n)$ works. Dividing both sides by $p^k$, we get $$g(p^k)=g(p^{k-1})+\frac{p-1}{p}.$$ As $k=0$ gives $1$, we know that $$g(p^k)=1+k\cdot\frac{p-1}{p}.$$ It is now clear that $x=2^{2a-2}$ satisfies the requested condition.
c) First, it is clear that $g(n)=1$ only at $n=1$, so $a=1$ is unique. We can express any positive integer $x>1$ as $x=\prod{p_i}^{e_i}$, and therefore, $$g(x)=g\left(\prod{p_i}^{e_i}\right)=\prod g\left({p_i}^{e_i}\right)=\prod\left(1+e_i\cdot\frac{p_i-1}{p_i}\right).$$ Note that if we simply set $e_i=p_i$ for a certain $i$, we would be multiplying by $p_i$. Therefore, if we had $a$ be any number that is not a power of $2$, we can take any prime factor of $a$ that is not $2$, set that as $p_i$ and $e_i$, and let $p_1=2$ do the rest of the work, giving us an unwanted second way to construct $g(x)=a$. However, if $a$ was a power of $2$, assume FTSOC that there is this other $y$ such that $g(y)=a$ and $y$ is not a power of $2$. We can see that if we were to take the product besides the term for $p_i=2$, the numerators are all odd, and the denominators are all odd. The numerator is clearly greater than the denominator, and that part of the numerator clearly cannot be cancelled out from the $p_i=2$ term. Therefore, $g(x)$ would have some factor that is not a power of $2$ that the denominator cannot cancel out, meaning that it does not evaluate to a power of $2$, contradiction. Therefore, our answer is $$\boxed{a=2^k,k\in\mathbb{Z}^{\ge0}}.$$

We claim that if $c\equiv a\pmod m$ and $c\equiv b\pmod n$ then $\gcd(c,mn)=\gcd(a,m)\gcd(b,n).$ Since $\gcd(a,m)=\gcd(c,m)$ and $\gcd(b,n)=\gcd(c,n)$ by the euclidean algorithm, and $\gcd(c,m)\gcd(c,n)=\gcd(c,mn)$ so our claim holds. By CRT, there is a bijection between $(a,b)$ and $c$ so $f(mn)=f(m)f(n)$

Note that $f(1)=1$ and $f(2^{k+1})=2f(2^k)+2^k$, so by induction, $f(4^k)=(k+1)4^k.$

We claim that the answer is powers of two. Suppose $p\mid a$ where $p$ is an odd prime. Already $4^{a-1}$ is a solution, but also let $a=pq,$ and we have $f(4^{q-1})=q.$

We claim that \[f\left(27^{\frac{p-1}{2}}\right)=p\cdot \left(27^{\frac{p-1}{2}}\right).\] We can do this in much the same way as $f(4^k)=(k+1)4^k$ and since $f(m)f(n)=f(mn)$ we're done, as we have found another solution to the equation.

[b]a)[/b] Note that every number $k$ from $1$ to $mn$ has a unique pair $(x,y)$ such that $k\equiv x\pmod m$ and $k\equiv y\pmod n$; this means that there is a bijection between values of $\gcd{(k,mn)}$ and values formed by multiplying $f(m)f(n)$ into $mn$ different summands.

[b]b)[/b] Note that
\[f(p^k)=p^k(1)+p^{k-1}(p-1)+p^{k-2}(p^2-p)+\dots+p(p^{k-1}-p^{k-2})+1(p^k-p^{k-1})=p^k\left(k\left(\frac{p-1}{p}\right)+1\right)\]
and so evidently $2^{2a-2}$ is a solution.

[b]c)[/b] Set $g(p,k)=k\left(\frac{p-1}{p}\right)+1$ and note that if $x=p_1^{e_1}\cdot p_2^{e_2}\cdots p_n^{e_n}$ then we have
\[g(p_1,e_1)g(p_2,e_2)\dots g(p_n,e_n)=a.\]
From part b) we know that $2^{2a-2}$ is a solution. Now suppose that $a=2^ko$ where $o>1$ is odd. Then set $x=2^{2^{k+1}-2}\cdot 3^{\tfrac{3o-3}{2}}$. This gives another solution. Then $a$ is a power of $2$. Then all powers of $2$ in $g(p_1,e_1)g(p_2,e_2)\dots g(p_n,e_n)$ must come from $g(2,2^{k+1}-2)$ where $a=2^k$ implying that $2^{2a-2}$ is the only solution so the answer is all powers of $2$. $\blacksquare$

If $d$ divides $n$, then the number of times $d$ shows up in $f(n)$ is the number of $q$ with $1\leq q\leq \tfrac{n}{d}$ such that $dq\leq n$ and $\gcd(q,n)=1$. This is the number of units $\mod \tfrac{n}{d}$, or $\phi(\tfrac{n}{d})$. So if $d\mid mn$, $d$ shows up in $f(mn)$ exactly $\phi(\tfrac{nm}{d})$ times. If $e\mid d$ shows up in $f(n)$ and $\tfrac{d}{e}$ shows up in $f(m)$, then $e\mid n$ and $\tfrac{d}{e}\mid m$, so the number of times $e$ shows up in $f(n)$ is $\phi(\tfrac{n}{e})$ and the number of times $\tfrac{d}{e}$ shows up in $f(n)$ is $\phi(\tfrac{m}{d/e})$. Moreover, we know that $e=\gcd(n,d)$ and $\tfrac{d}{e}=\gcd(m,d)$, so the only values in $f(n)$ and $f(m)$ that multiply to $d$ are $\gcd(n,d)$ and $\gcd(m,d)$, respectively. So the number of terms in $f(n)\cdot f(m)$ that are equal to $d$ is $\phi(\tfrac{n}{e})\cdot \phi(\tfrac{m}{d/e})=\phi(\tfrac{nm}{d})$, since $m,n$ are coprime. Therefore for each $d\mid nm$, it shows up the same number of times in $f(mn)$ and $f(m)f(n)$. The number of terms in $f(mn)$ is the same as the number of terms in $f(m)f(n)$, thus if $\gcd(m,n)=1$, then $f(mn)=f(m)f(n)$, which solves part (a).

    We now find a formula for $f(n)$. If $k=\ell p^d$ where $1\leq \ell \leq p^{e-d}$, and $\gcd(\ell,p)=1$, then $\gcd(k,p^e)=p^d$, and there are $\phi(p^{e-d})$ values of $\ell $ satisfying this. Since $\gcd(k,p^e)\neq 1\implies p\mid k$, we know that
    \begin{align*}
        f(p^e)&=1\cdot \phi(p^e)+\sum_{d=1}^{e-1} p^d\cdot \phi(p^{e-d})+p^e\cdot \phi(1) \\
        &= \sum_{d=0}^e p^d\cdot \phi(p^{e-d}) \\
        &= \sum_{c\mid p^e} \phi(c)\cdot \frac{p^e}{c}.
    \end{align*}
Then, evaluating $f(p^e)$ for $e\geq 1$ and prime $p$, we get $f(p^e)=p^{e-1}(p(e+1)-e)$. Now note that if $e=pr$ for some positive integer $r$, we get $f(p^{pr})=p^{pr}(r(p-1)+1)$. Setting $p=2$ and $e=2e_1$, we have $f(2^{2e_1})=2^{2e_1}(e_1+1)$. Thus $f(2^{2(a-1)})=a2^{2(a-1)}$ for all $a\in \mathbb{N}$. This solves part (b).

    Note that if $e=\tfrac{p^b-1}{p-1}\cdot p$, then $f(p^e)=p^e\cdot p^b$ for any $b\geq 1$, using our formula for $f(x)$. Since $f$ is multiplicative, if we let $a=p_1^{e_1}\cdots p_k^{e_k}$ be the prime factorization of $a$, then a solution to $f(x)=ax$ is $x=\Pi_{i=1}^k p_i^{d_i}$ where $d_i=\tfrac{p(p^{e_i}-1)}{p-1}$ for all $1\leq i\leq k$. If $a$ is not a power of $2$, then this value of $x$ is not $2^{2(a-1)}$, since it contains prime factors other than $2$.

    Suppose $a=2^e$, for some non-negative integer $e$, and suppose $x=p_1^{e_1}\cdots p_k^{e_k}$ is the prime factorization of a solution to $f(x)=2^e x$. Then $$f(x)=\prod_{i=1}^k p_i^{e_i-1}(p_i(e_i+1)-e_i)=2^e(p_1^{e_1}\cdots p_k^{e_k}),$$ so $$\prod_{i=1}^k (p_i(e_i+1)-e_i)=2^e(p_1\cdots p_k).$$ Each $p_i(e_i+1)-e_i$ that has $p_i\neq 2$ is odd, since $p_i>2$ by casework on the parity of $e_i$. Thus the only way $2^e\mid \Pi_{i=1}^k (p_i(e_i+1)-e_i)$ is if there is a $p_i$ that is $2$ (WLOG let $p_1=2$) and $2^e\mid 2(e_1+1)-e_1$. This implies that $2^e\mid e_1+2$ so $e_1+2=2^e\cdot m$ for some $m\in \mathbb{N}$. Also, since $p_1=2$, we have $2\mid m$, so $m=2m_1$ for some $m_1\in \mathbb{N}$. Then we have $$\prod_{i=2}^k ((p_i-1)e_i+p_i)\cdot m_1=p_2\cdots p_k.$$ For each $i$, if $e_i>0$, then $(p_i-1)e_i>0$, which means that the LHS of the above equation is larger than the RHS, contradiction. Therefore for all $i$, we must have $e_i=0$. But then the solution $x=p_1^{e_1}\cdots p_k^{e_k}$ to $f(x)=2^e x$ must be of the form $2^{e_1}$, and as we have seen, $e_1=2^em-2$ for some $m$. We get $$f(2^{2^em-2})=2^{2^e-3}(2(2^em-1)-(2^em-2))=2^e\cdot 2^{2^em-2},$$ which simplifies to $m=2$. Thus $x=2^{2(2^e-2)}$, meaning that $f(x)=ax$ has a unique solution if and only if $a=2^e$ for some $e\in \mathbb{Z}_{\geq 0}$.

By counting over values of $\gcd(k,n)$, we express $f$ as
$$f(n)=\sum_{d \mid n} d\cdot\phi(n/d)=\mathrm{id} * \phi$$
where $*$ denotes the usual Dirichlet convolution. From this, multiplicativity is immediate. For $p$ prime, we can then compute
$$f(p^k)=(k+1)p^k-kp^{k-1}.$$
We now present two methods of generating a solution to $f(x)=ax$ for any $a$ not a power of two, and prove that $f(x)=ax$ has a unique solution when $a$ is a power of two. For any $a$, we can simply take $x=2^{2a-2}$, in which case
$$f(x)=f(2^{2a-2})=(2a-1)2^{2a-2}-(2a-2)2^{2a-3}=2^{2a-2}(2a-1-(a-1))=a2^{2a-2},$$
as desired. On the other hand, from our formula for $f(p^k)$, fixing $p$ and setting $k=\tfrac{p^{a+1}-p}{p-1}$, we obtain
$$f(p^k)=\frac{p^{a+1}-1}{p-1}\cdot p^k-\frac{p^{a+1}-p}{p-1}p^{k-1}=p^k\cdot \frac{p^{a+1}-p^a}{p-1}=p^ap^k.$$
By multiplicativity, for some $a=p_1^{e_1}\ldots p_k^{e_k}$, taking
$$x=\prod_{i=1}^k p_i^{\frac{p^{e_i+1}-p}{p-1}}$$
satisfies $f(x)=ax$. Moreover, if $a$ is not a power of two, then this value of $x$ is different from $2^{2a-2}$ as it is divisible by some odd prime.
If $a$ is a power of two, note that $x=2^{2a-2}$ can easily be seen to be the only power of two that yields $f(x)=ax$. Hence suppose we have some $x$ with $f(x)=ax$ that's a non-power of two, and suppose $\nu_2(x)=k$. For any $p>2$, $f(p^k)$ is odd, hence $\nu_2(f(x))=\nu_2(f(2^k))$. If $k$ is odd, then $\nu_2(f(2^k))=k-1$, but $\nu_2(ax)\geq k$: impossible. If $k$ is even, then from our previous results $\nu_2(f(2^k))=k+\nu_2(\tfrac{k}{2}+1)$, hence $\nu_2(\tfrac{k}{2}+1)=\nu_2(a) \implies k\geq 2a-2$, since $a$ is a power of two. Further, since $f(p^k)>p^k$ for any odd $p$, and $\tfrac{f(2^k)}{2^k}=k/2+1$ is increasing in $k$, we have
$$\frac{f(x)}{x}=\frac{f(2^k)}{2^k}\cdot \frac{f(x/2^k)}{x/2^k}\geq \frac{f(2^{2a-2})}{2^{2a-2}}\cdot \frac{f(x/2^k)}{x/2^k}>\frac{f(2^{2a-2})}{2^{2a-2}}=a,$$
which is a contradiction. Hence $x=2^{2a-2}$ is the unique solution to $f(x)=ax$, as desired. $\blacksquare$

(a) Note that if we are going from 1 to $mn$, by CRT, each possible pair of residues mod $m$ and $n$ shows up exactly once. If $m$ and $n$ are relatively prime $x\equiv a\pmod m$ and $x\equiv b\pmod n$, then $\gcd(x,mn)=\gcd(a,m)\cdot \gcd(b,n)$, so therefore, we have $$f(mn)=\sum_{i=0}^{m-1}\sum_{j=0}^{n-1}(\gcd(i,m)\cdot \gcd(j,n))$$ $$=\sum_{i=0}^{m-1}(\gcd(i,m)\cdot \sum_{j=0}^{n-1}(\gcd(j,n)))$$ $$=\sum_{i=0}^{m-1}(\gcd(i,m)\cdot f(n))$$ $$=f(n)\sum_{i=0}^{m-1}\gcd(i,m)$$ $$=f(m)f(n).$$

(b) We first claim that if $p$ is a prime and $k$ a positive integer, then $$f(p^k)=(k+1)p^k -kp^{k-1}.$$ We will use induction. For $k=1$, we obviously have $f(p)=2p-1.$ Now consider $f(p^{k+1})$ given $f(p^k)$. Note that the values in the summation for $f(p^{k+1})$ is almost the same as $f(p^k)$ repeated $k$ times, but with the final term being $p^{k+1}$ rather than $p^k$ again. Therefore, we have the recurrence $$f(p^{k+1})=pf(p^k)+p^{k+1}-p^k,$$ from which it follows by induction that $$f(p^k)=(k+1)p^k -kp^{k-1}.$$

In particular, we have $$f(2^k)=(k+1)2^k -k\cdot 2^{k-1}$$ $$\frac{f(2^k)}{2^k}=k+1-\frac{k}{2}=\frac{k}{2}+1,$$ so we can obtain any positive integer value of $f(x)/x$ by plugging in a power of 4 (possibly 1).\\

(c) We claim that the answer is all powers of 2. If $a$ is not a power of 2. consider the number $2^{2n}\cdot 3^{3m}.$ Since $f$ is multiplicative, and again using our prime power formula from earlier, we can obtain that $$\frac{f(2^{2n}\cdot 3^{3m})}{2^{2n}\cdot 3^{3m}}=(n+1)(2m+1).$$ If $a$ is not a power of 2, we can make $m$ a positive integer to make the required odd factor, then adjust $n$ to get all the remaining factors of 2, so there is a way to achieve $a$ using at least one factor of 3. However, we also know from earlier that we can achieve any value of $a$ using only factors of 2, so there is more than one way to achieve $a$ if $a$ is not a power of 2.\\

To show that there is only one way to achieve powers of 2, consider the following. Let $n$ be a positive integer. Then, since the function is multplicative, we have $$\frac{f(n)}{n}=(\frac{1}{2}v_2(n)+1)(\frac{2}{3}v_3(n)+1)(\frac{4}{5}v_5(n)+1)(\frac{6}{7}v_7(n)+1)\cdots$$ Consider how we could get a power of 2 from this function. Note that $$(\frac{2}{3}v_3(n)+1)(\frac{4}{5}v_5(n)+1)(\frac{6}{7}v_7(n)+1)\cdots\geq 1.$$ Moreover, if $n$ has any prime factor other than 2, i.e. if any of $v_3(n),v_5(n),$ etc. are positive, then $$(\frac{2}{3}v_3(n)+1)(\frac{4}{5}v_5(n)+1)(\frac{6}{7}v_7(n)+1)\cdots$$ will be greater than 1 and have an odd numerator, which will prevent the expression from being a power of 2. Therefore, the only way for $f(n)/n$ to be a power of 2 is if $n$ itself is a power of 2, so if $a$ is a power of 2, there is only one way to achieve it, so we are done.

Can someone please tell me where I am going wrong in the proof for $\textbf{part (b)}$ T__T :( :noo: .

We firstly have that $\varphi * 1=\operatorname{id}\iff\varphi=\operatorname{id}*\mu$. The $d(n)$ is the number of divisors of $n$, $\sigma(n)$ is the sum of divisors of $n$, $\varphi(n)$ is the Euler Totient Function, $\operatorname{id}$ is the identity function and $\mu(n)$ is the Mobius Function.

So after getting $f(n)=\varphi *\operatorname{id}$, we can get the following $f(n)=(\operatorname{id}*\mu)*\operatorname{id}=\operatorname{id}*(\operatorname{id}*\mu)=(\operatorname{id}*\operatorname{id})*\mu=F*\mu$ where $F(n)=nd(n)$. We can use Mobius Inversion again to get $f*1=F$.

Now for $\textbf{part (b)}$, [hide=fakesolve part]we put $f(n)=an$ which along with our last claim gives,\[nd(n)=F(n)=\sum_{d\mid n} f(d)=\sum_{d\mid n} ad =a\sum_{d\mid n}d=a\sigma(n),\]which finally gives that we need to find $\dfrac{f(n)}n=a=\dfrac{n\cdot d(n)}{\sigma(n)}$.

This however is just $\dfrac{f(n)}{n}$ but according to the definition of the other solutions for $\dfrac{f(n)}{n}$, my one completely contradicts the form that the others have arrived at which must mean that I am going wrong somewhere :maybe: . Can someone please help me :sob: .

Like for say vEnhance's solution has $g(n)=\dfrac{f(n)}n$ and it shows that $g(p^e)=e + 1 - \frac ep$ but my one gives $g(p^e)=\dfrac{p^e(e+1)(p-1)}{p^{e+1}-1}$ which aren't alike at all.[/hide].

Thanks to @a_n for pointing out where I went wrong :-D :blush: .

We first consider for how many occurrences of $k$ for which we have $\gcd(k,n)=d$, where $d$ is a divisor of $n$. We must have $d\mid k$, and $\gcd(\frac{k}{d},\frac{n}{d})=1$. That is, $\frac{k}{d}$ is coprime to $\frac{n}{d}$. So, there are $\phi(\frac{n}{d})$ occurrences of $k$. Our sum then becomes
\[f(n)=\sum_{k=1}^n\gcd(k,n)=\sum_{d\mid n}d\phi(\frac{n}{d})=(I*\phi)(n)\]
where $I$ denotes the identity function. This immediately gives us (a), since $f$ is multiplicative as a consequence of being the Dirichlet convolution of two multiplicative functions.

For a prime $p$, we have $f(p^k)=\sum_{e=0}^{k}p^e\phi(p^{k-e})=p^0(p-1)p^{k-1}+p^1(p-1)p^{k-2}+\dots + p^{k-1}(p-1)p^0+p^k=k(p-1)p^{k-1}+p^k$. Now consider the function $g(n)=\frac{f(n)}{n}$, which is obviously multiplicative. We have $g(p^k)=\frac{k(p-1)p^{k-1}+p^k}{p^k}=k\frac{p-1}{p}+1$. For each positive integer $a$, we have $g(2^{2(a-1)})=2(a-1)\frac{1}{2}+1=a$. This proves (b).

We claim for (c) that the answer is all $a$ which are a power of $2$. Suppose that $a$ is not a power of $2$. Then $a$ has an odd factor greater than $1$ and can be written as $c(2d+1)$ for some positive integers $c$ and $d$. We have already that $g(2^{2(a-1)})=a$, and here we have also $g(2^{2(c-1)}3^{3d})=g(2^{2(c-1)})g(3^{3d})=c(3d\frac{2}{3}+1)=c(2d+1)=a$.

So, for all $a$ which are not a power of $2$, $g(n)=a$ has more than one solution. We now show that when $a$ is some power of $2$ $2^r$, $g(n)=a$ has only one solution. Observe that for all odd primes $p$, $f(p^k)=k(p-1)p^{k-1}+p^k$ is odd. Consider $f(n)=2^ln$ and let $s=\nu_2(n)$. $\nu_2(f(n))=\nu_2(2^ln)=l+s$. Thus, we must have that $2^{l+s}$ divides $g(2^s)=s2^{s-1}+2^s$. So, $s2^{s-1}+2^s\geq 2^{l+s}$. Dividing both sides by $2^s$ we get $\frac{s}{2}+1\geq 2^l$ and $s\geq 2(2^l-1)$.

Considering $2^l=g(n)=g(2^s)g(3^{\nu_3(n)})g(5^{\nu_5(n)})\dots$, we note that $g(p^k)=k\frac{p-1}{p}+1$ is increasing in $k$ for all primes $p$. We have also that $g(2^{2(2^l-1)})=2^l$. We deduce that $s=l$ and $\nu_p(n)=0$ for all odd primes $p$. Thus, $g(n)=2^l$ has only the solution $n=2^{2(2^l-1)}$, which proves our answer for (c).

Nice!

(a) Note $f = id * \phi$, and since $id$ and  $\phi$ is multiplicative, $f$ is multiplicative too.

(b) Note that $f(p^k) = p^{k-1}((k+1)p-k)$, so we want, if possible, $(k+1)p-k = ap \implies p =1+ \frac{a-1}{k+1-a}$.  Thus, we find that $k = 2a-2, p=2$ works and $f(2^{2a-2}) = a2^{2a-2}$.

(c)
Claim: All $a$ which are powers of 2(greater than 1) have exactly one solution
Proof: This is because for $f(p^k) = p^{k-1}((k+1)p-k)$, we have that for $p>2$, $(k+1)p-k$ is odd.  Since the function is multiplicative, in order for $a$ to be a power of 2, we have to have $x$ is a power of two in which there is a distinct soltuion.
Claim: There is more than one solution when $a$ is odd
Proof: Note that $f(3^{3k}) = (2k+1)3^{3k}$, proving our claim.
Therefore, all even $a$ that are not powers of 2 can be created in two ways (either just using power of 2 or using power of 2 and power of 3).

Thus, there is a unique solution for $a$ when $a$ is a power of 2 greater than 1.

[b][color=pink]Claim:[/color][/b] $f = \phi \ast id$ and is thus multiplicative.
[i]Proof.[/i] Note that \[ f(n) = \sum_{k=1}^n \gcd(k,n) = \sum_{d \mid n} \phi\left(\frac{n}{d}\right) \cdot d. \] $\blacksquare$
We thus compute $f(p^e)$ \[ f(p^e) = \sum_{d=0}^e \phi(p^d) \cdot p^{e-d} = p^e + \sum_{d=1}^e p^{d-1} \cdot (p-1) \cdot p^{e-d} = p^e + ep^{e-1} \cdot (p-1). \]
As such, \[ \frac{f(p^e)}{p^e} = 1 + e - \frac{e}{p}. \]
This generalizes for $n = p_1^{e_1}p_2^{e_2} \dots p_k^{e_k}$ as \[ \frac{f(n)}{n} = \frac{f(p_1^{e_1})}{p_1^{e_1}} \cdot  \frac{f(p_2^{e_2})}{p_2^{e_2}} \cdots \frac{f(p_k^{e_k})}{p_k^{e_k}} \]
For each integer $a$, we have that $f(2^{2(a-1)}) = a \cdot 2^{2(a-1)}$. If $a$ is a power of $2$, since $\frac{f(p^e)}{p^e}$ has an odd divisor of the numerator for $p \ne 2$, it only has one factor.
Else, we also another construction as \[ \frac{f(3^e)}{3^e} = \frac{2e + 3}{3} \] which allows constructing all odds other than $1$.

Note that since
\[ f(n) = \sum_{k-1}^n \gcd(k, n) = \sum_{d\mid n} d \cdot \phi \left(\frac{n}{d}\right) \]
Thus $f = id * \phi$ and thus $f$ is multiplicative. This completes part a. $\square$

Now, note that for any positive integer $k$,
\[ f(p^k) = p^k - p^{k-1} + p(p^{k-1} - p^{k-2}) + + p^2(p^{k-2} - p^{k-3}) + \dots + p^{k-1}(p-1) + p^k = k(p^k - p^{k-1}) + p^k = (kp+p-k)p^{k-1} \]
In particular, note that if $p^k \mid (kp+p-k)p^{k-1}$ then $k = jp$, so
\[ f(p^k) = (jp^2 + p - jp)p^{k-1} = (jp - j + 1)p^k = (j(p-1)+1)p^k \]
This means that if $p = 2$ and $k = 2j$, then
\[ \frac{f(2^{2j})}{2^{2j}} = j + 1 \]
Thus, then $f(x) = ax$ always has a solution, by taking $n$ a power of $4$ and choosing the appropriate exponent. This completes part b. $\square$

Note also that for an odd prime $p$ and $k = jp$, then
\[ f(p^k) = (kp+p-k)p^{k-1} = (jp+1-j)p^k = (j(p-1)+1) \]
Then this means that
\[ \frac{f(p^k)}{p^k} = j(p-1)+1 \]
This value is always odd when $p>3$. Now, since
\[ \frac{f(n)}{n} = \prod_{p\mid n} \frac{f(p^{\nu_p(n)})}{p^{\nu_p(n)}} \]
if $\frac{f(n)}{n} = a$ and $a$ is divisible by a prime other than $2$, then we could use either $n = 2^b$ or $n = 2^b3^c$, where the $3^c$ can take care of the largest odd divisor of $a$. Thus, then the only $a$ with a unique solution is when $a$ is a power of $2$. $\blacksquare$

We have \[ f(n) = \sum_{d\mid n} d \cdot \varphi \left( \frac nd \right) \] because there are exactly $\varphi \left( \frac nd \right )$ positive integers in $k\in [1,n]$ with $\gcd(k,n) = d$. Hence by the Dirichlet convulution, since identity and $\phi$ are multiplicative, we must have $f$ multiplicative.

Now, we also see that \[f(p^t) = \sum_{i=0}^t p^i \varphi (p^{t-i} )  = p^t + t(p-1) p^{t-1} \] for all positive integers $t$.

Hence $\frac{f(p^t)}{p^t} = t \cdot \frac{p-1}{p} + 1$. We want to show that for all $a$, there exists $x$ with $\frac{f(x)}{x}= a$.

Notice that if $p = 2$, then $\frac{f(2^t)}{2^t} = \frac{t}{2} + 1$. Choosing $t = 2(x-1)$ gives that $f(2^t) = x\cdot 2^t$.

Claim: The answer to part (c) is $a$ is a power of $2$.
Proof: For non powers of $2$, choose $x = 2^y \cdot 3^z$ so that $\frac{2z}{3} + 1 = m$ is the largest odd divisor of $a$ and $\frac y2 + 1 = \frac{a}{m}$. Now it remains to show that only one $x$ satisfies $\frac{f(x)}{x} = 2^e$ for each $e$. Notice that $\frac{f(p^t)}{p^t} \equiv 1\pmod q$ for any prime $q\mid p-1$. Hence $p-1$ cannot have any prime factors, so $p = 2$. But then $t = 2(2^e - 1)$, which means there is only one solution to $f(x) = 2(2^e - 1)$. $\square$


Let $r,s,t$ be roots of $x(x-2)(3x-7) = 0.$ Prove that $r,s,t$ are not negative and evaluate $\tan^{-1}{r} + \tan^{-1}{s} + \tan^{-1}{t}.$

Chigr wrote:[hide]well to get the roots of the equation we split each part so that each equals zero so:
x=0
x-2=0, x=2
3x-7=0,3x=7, x=7/3
r, s, and t are therefore 0, 2 and 7/3 which are all non-negative numbers

tan^-1(0)+tan^-1(2)+tan^-1(7/3)
~130.24[/hide]

I don't think u can use a calculator and a decimal approximation

Could you please explain what arctans are, no one at my school has ever mentioned them, which isn't surprising since we only know tan, cos and sin and haven't even done csc or sec etc.


inequalities
complex numbers
algebra unsolved
algebra

Prove that the complex numbers a1,a2,..,an (n>=2) satisfy the following inequality :
| a1+a2*z+...+an*z^(n-1)|<=|a1+a2+...+an| , z from C with |z|<=1 if and only if there exist the  pozitive real numbers b2,b3,..,bn (<>0)   so that a1=b2*a2=...=bn*an.

Please help me! Thank you!

Louis.

I am still looking for answers...  :?
Hope someone will give me one.... :D

[quote="Myth"]I don't believe Lagrangia said that! Obviously, the statement is wrong.[/quote]
Yes, it is wrong. Let $b_{1},b_{2},...,b_{n}$ complex numbers, suth that $arg(b_{1}),arg(b_{2}),...,arg(b_{n})$ don't belongs to arithmetic progression.
Let $\phi_{0}$ give maximum $\sum_{i}b_{k}exp(ik\phi)$. Then $a_{k}=b_{k}exp(ik\phi_{0})$ give contr example your statement.


ARML
geometry
calculus
integration
trigonometry
perpendicular bisector
power of a point

1.
In right triangle $ABC$, $\angle C = 90^\circ$, $c=3$, and $a+b=\sqrt{17}$.  Compute the area of the triangle.

2.
If $(a+b) : (b+c) : (c+a) = 6 : 7 : 8$, and $a+b+c=14$, compute the value of $c$.

3.
Let $z$ be a root of $z^5-1=0$, with $z \not= 1$.  Compute the value of

\[z^{15} + z^{16} + z^{17} + \cdots + z^{50}. \]

4.
A regular polygon of $2n$ sides is inscribed in a circle of radius 5.  The sum of the squares of the distances from any point on the circle to the vertices of the polygon is 600.  Compute the value of the integer $n$.

5.
The circles, each of radius 5, lie in perpendicular planes.  They intersect in two points, $X$ and $Y$, thus determining a common chord $\overline{XY}$.  If $XY = 8$, compute the distance between the centers of the circles.

6.
How many positive integers less than 54 are equal to the product of their proper divisors?

7.
Find all positive integers $n$, less than 17, for which $n! + (n+1)! + (n+2)!$ is an integral multiple of 49.

8.
In triangle $ABC$, $a \ge b \ge c$.  If $\displaystyle \frac{a^3+b^3+c^3}{\sin^3{A} + \sin^3{B} + \sin^3{C}} = 7$, compute the maximum possible value of $a$.

[hide="3"]
The root is $z=e^{\pi i\frac{k}{5}}$ where $k\in\{1,2,3,4\}$
Note that $z^{1},z^{2},z^{3},...$ will end up repeating with period 5 using the each root of $z^{5}-1=0$ Since the coefficient of $x^{4}$ is 0, the sum of those roots is $0$. All sets of five within the series cancel out to $0$. The only term not within a set is $z^{50}=\boxed{1}$
[/hide]

[hide="7"]
$n!+(n+1)!+(n+2)!=n!+(n+1)n!+(n+1)(n+2)n!=(1+n+1+(n+1)(n+2))n!=(2+n+n^{2}+3n+2)n!=(n^{2}+4n+4)n!=(n+2)^{2}n!$

So either 7 divides $n+2$, in which case $n=5\text{ or }12$ or 49 divides $n!$, in which case $n\ge14$

So the possible solutions are $\boxed{5,12,14,15,16}$
[/hide]

[/hide]

[hide="4"]
Pick a point $ P$ on the circle.

Choose a vertex $ V$ of the n-gon and the opposite vertex $ V'$. Then $ VV'$ is a diameter, so $ \angle VPV'$ is a right angle. Then $ PV^2\plus{}PV'^2\equal{}(2\cdot 5)^2\equal{}100$

There are $ n$ pairs of vertices, so summing all $ n$ pairs:

$ 100n\equal{}600$
$ n\equal{}6$
[/hide]

[hide="8"]
Use the law of sines.

$ 7\equal{}\frac{a^3\plus{}b^3\plus{}c^3}{\sin^3 A\plus{}\sin^3 B\plus{}\sin^3 C}\equal{}\frac{a^3\plus{}b^3\plus{}c^3}{\sin^3 A\plus{}\frac{b^3}{a^3}\sin^3 A\plus{}\frac{c^3}{a^3}\sin^3 A}\equal{}$
$ \frac{a^3\plus{}b^3\plus{}c^3}{\sin^3 A}\frac{a^3}{a^3\plus{}b^3\plus{}c^3}\equal{}\frac{a^3}{\sin^3 A}$

$ a^3\equal{}7\sin^3 A\le 7$
$ a\le\sqrt[3]{7}$
[/hide]

[hide="2"]$ a \plus{} b \plus{} c \equal{} 14$
$ 2a \plus{} 2b \plus{} 2c \equal{} 28 \equal{} (a \plus{} c) \plus{} (b \plus{} c) \plus{} (a \plus{} b) \equal{} 6k \plus{} 7k \plus{} 8k \equal{} 21k$
$ 28 \equal{} 21k$
$ k \equal{} \frac {4}{3}$
$ a \plus{} b \equal{} 6k \equal{} 6(\frac {4}{3}) \equal{} 8$
$ c \equal{} 14 \minus{} (a \plus{} b) \equal{} 14 \minus{} 8 \equal{} 6$[/hide]
[hide="5"] The chord's perpendicular bisector is a diameter of the circle, so the chord is divided into 4 and 4.  Then, using the power of a point theorem,
$ 4\times{4} \equal{} x\times{(10 \minus{} x)}$
$ 16 \equal{} 10x \minus{} x^2$
$ x^2 \minus{} 10x \plus{} 16 \equal{} 0$
$ (x \minus{} 8)(x \minus{} 2) \equal{} 0$
$ x \equal{} 8, 2$
Notice that these are not 2 unique solutions; they are essentially the same.  So the distance from the center of the circle to the chord is 5-2=3.  The centers of the circles and the center of the chord make an isosceles right triangle with side length 3.  So the hypotenuse, and the distance between the centers of the circles, is $ 3\sqrt {2}$.[/hide]


eh here goes:
in a sequence of numbers the sum is Sn=5n^2+n
find a1 a2 a3 and an.

it would be nice for you to show some work but its for homework so do what you can.

The sum of an arithmetic sequence can be defined as follows, (where the number in the parentheses represents the term number)

S(n) = 1/2 * [a(1)+a(n)]

I shall prove this, where d is the common difference.

S(n) = a(1)+ [a(1) + d] + ... + [a(1) + (n - 2)d] + [a(1) + (n - 1)d]
S(n) = [a(1) + (n - 1)d] + [a(1) + (n - 2)d] + ... + [a(1) + d] +a1

Grouping the terms, we have

2S(n) = {a1 + [a(1) + (n - 1)d]} + {a(1) + [a(1) + (n - 1)d]} + ...
                +{a(1) + [a(1) + (n - 1)d]} + {a1 + [a1 + (n - 1)d]}

        =  [a1 + an] + [a1 + an] +... + [a1 + an] + [a1 + an]

        = n[a(1) + a(n)]

So,

S(n) = n/2 [a(1)+a(n)]

Now, referring to your post, we can substitute in 5n^2+n.

5n^2+n=n/2 [a(1)+a(n)]
5n^2+n=n/2 [a(1)+a(1)+(n-1)d]
10n^2+2n=n [2a(1) + (n-1)d]
n(10n+2) = n [2a(1) + (n-1)d]

Sine n cannot be zero, we divide by n.

10n+2 = 2a(1) + (n-1)d
10n+2-(n-1)d = 2a(1)
[10n+2-(n-1)d]/2 = a(1)

Ok, now I don't know what I'm doing. Since there are many, many different solutions to this equation, I don't think if your sequence is unique.

Liyis way is a much easier method. But that way does also work - just remember that the last equation you came up with is not to be solved for d, n, and a(1).. but its true for ALL n.

wow.. that looks difficult.  but where is the class for the hw?
anyway, hi dyl


It's wrong if $ a<d$ .^^  .Do you think :$ LHS\equal{}(1\minus{}a)(1\minus{}b)(1\minus{}c)(1\minus{}d)$?

By  CBS  $ \sqrt{(a^2\plus{}b^2)(c^2\plus{}d^2)} \geq ac\plus{}bd$  -->Ok  easy

we have
 2(1-a)(1-d)= $ (a\plus{}d\minus{}1)^2  \plus{}b^2\plus{}c^2  \geq b^2\plus{}c^2$ --->  DONE   hahahahahaha :P


geometry
trigonometry
inradius
trig identities
Law of Cosines
area of a triangle
Heron\u0027s formula

[quote="neoangel06"]Okay I have this equation and I need help with it.

There is this Triangle. Point A, B, and C. The length of AB is 240 km, BC is 330 km, and CA is 270 Km. There is a circle inside which fits perfectly. What is the area of the circle in square km's?
I have tried and tried to come up with a solution, but I cannot find one. Somebody please help![/quote]

Here's one method, dunno if you just wanted hints or not.

[hide]
Let X, Y, and Z be the points where the circle intersects the sides AB, BC, and AC, respectively.

Then AX = AZ, BX = BY, CY = CZ.
Also, AX + BX = AB = 240, AZ + CZ = AC = 270, BY + CY = BC = 330.

Solve for the lengths:
AX = 90, BX = 150, CY = 180.  You can check these values by adding them up in pairs to get the original lengths.

Use Law of Cosines to get an angle.  Use simple trig with half-angle to get a radius.  Compute area.

There might be an easier way to get the radius once you know the lengths by some clever manipulation, but the trig method is straightforward for me.
[/hide]

Calculate the triangle semiperimeter $s = \frac{a+b+c}{2}$ and area by Heron's formula $\triangle = \sqrt{s(s-a)(s-b)(s-c)}.$ The inradius is $r = \frac{\triangle}{s}.$


Prove that if a unit square is partitioned into two sets, then the diameter of one of the sets is not less than $ \sqrt5/2$. Show that no larger number will do. (The diameter of a set is the least upper bound of the distances between all pairs of points in the set.)

Let us color points from the first set green and from the second set red.

WLOG let $A$ be green, and suppose the problem statement is false. Then $B,D$ must be red, because $AB,AD\geq  \frac{\sqrt{5}}{2}$, and this implies that $C$ is green, because $BC=\frac{\sqrt{5}}{2}$.

But from this we can conclude that $E$ can't be red because $ED=\frac{\sqrt{5}}{2}$, and it can't be green because $EC=\frac{\sqrt{5}}{2}$. Contradiction, as desired.

($\frac{\sqrt{5}}{2}$ is the best possible because we can color the upper half of the square green and the bottom half red, and then both diameters are $\frac{\sqrt{5}}{2}$)


http://www.artofproblemsolving.com/Wiki/index.php/2006_AMC_10B_Problems/Problem_23

[asy]unitsize(2cm);
defaultpen(.8);

pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);
pair F = intersectionpoint( A--D, B--Ep );

draw( A -- B -- C -- cycle );
draw( A -- D );
draw( B -- Ep );
filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );

label("$7$",(1.45,0.15));
label("$7$",(2.2,0.45));
label("$3$",(0.45,0.35));

draw( C -- F, dashed );

label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
label("$D$",D,NE);
label("$E$",Ep,NW);
label("$F$",F,S);

label("$x$",(1,1));
label("$y$",(1.6,1));[/asy]

So in the solution, it reads "Since triangles $ EFA$  and $ BFA$  share an altitude from $ A$ and their respective areas are in the ratio 3:7, their bases must be in the same ratio, hence DF:CF = 3:7."

I don't understand how that works out. I thought the bases were EF and FB.

I think this is just a typo , those triangles share the altitude from B, the rest is easy the proportion of areas comes from some triangle similarities

Edited in the wiki, I think there was also a typo on the next line so I changed it as well   :)

Can someone please have a look at it so it makes sense?

Thanks


$a^2+(S-x_k)^2\geq 2a(n-1)(\frac{x_1 x_2 .....x_n}{x_k})^{\frac{1}{n-1}}$ and the result follows from Muirhead  :)  :)


If the real polynomial $P(x)= x^n+a_1x^{n-1}+...+a_{n-1}x+a_n$ has  only positive roots, we have the inequality
  $na_n \leq a_1a_{n-1}$ :)

[quote="t0rajir0u"]I think the left side should be $n^2 a_n$.  ;)[/quote]
  What made you thinks $n^2$. May I know it? In fact, this problem is very easy.  :)

[hide="Counterexample"]
Consider $P(x) = x^3 - 4x^2 + 5x - 2$.  It has positive roots $1, 1, 2$.

$a_1 = -4$
$a_{n-1} = a_2 = 5$
$a_n = a_3 = -2$

$na_n = -6$
$a_1 a_2 = -20$

But $-6 > -20$.
[/hide]

I think what you want is all negative roots, and I reiterate the LHS should be $n^2$.

[hide="In fact..."] The problem statement is, of course, AM-HM in disguise, although written incorrectly. [/hide]


Find a sequence of maximal length consisting of non-zero integers in which the sum of any seven consecutive terms is positive and that of any eleven consecutive terms is negative.

Suppose that this sequence has 17 terms. Then we will have that
$a_1+a_2+a_3+\cdot\cdot\cdot +a_{11}<0$
$a_2+a_3+a_4+\cdot\cdot\cdot +a_{12}<0$
$a_3+a_4++a_5\cdot\cdot\cdot +a_{13}<0$
$\cdot\cdot\cdot$
$a_7+a_8+a_9+\cdot\cdot\cdot +a_{17}<0$
And if we add this seven inequalities we will have a sum that is negative, but each column is positive, so the contradiction shows that the terms of this sequence are at most 16.
And here is an example $-5, -5, 13, -5, -5, -5, 13, -5, -5, 13, -5, -5, -5, 13, -5, -5$

It's weird that this problem is almost the same as IMO 1977: http://www.mathlinks.ro/viewtopic.php?t=61033

[quote="Beat"]Suppose that this sequence has 17 terms. Then we will have that
$ a_1 \plus{} a_2 \plus{} a_3 \plus{} \cdot\cdot\cdot \plus{} a_{11} < 0$
$ a_2 \plus{} a_3 \plus{} a_4 \plus{} \cdot\cdot\cdot \plus{} a_{12} < 0$
$ a_3 \plus{} a_4 \plus{} \plus{} a_5\cdot\cdot\cdot \plus{} a_{13} < 0$
$ \cdot\cdot\cdot$
$ a_7 \plus{} a_8 \plus{} a_9 \plus{} \cdot\cdot\cdot \plus{} a_{17} < 0$
And if we add this seven inequalities we will have a sum that is negative, but each column is positive, so the contradiction shows that the terms of this sequence are at most 16.
And here is an example $ \minus{} 5, \minus{} 5, 13, \minus{} 5, \minus{} 5, \minus{} 5, 13, \minus{} 5, \minus{} 5, 13, \minus{} 5, \minus{} 5, \minus{} 5, 13, \minus{} 5, \minus{} 5$[/quote]

Quite simple and nice !
Congrats
:)


let $ a$ , $ b$ , $ c$ $ \geq 0$ , prove that :
1)
$ \frac {ab}{(a \plus{} c)(c \plus{} b)} \plus{} \frac {bc}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {ca}{(c \plus{} b)(c \plus{} a)} \geq \frac {3}{4}$

2) if $ a \plus{} b \plus{} c \equal{} 1$ :
$ \frac {a}{(b \plus{} 1)(1 \plus{} c)} \plus{} \frac {b}{(c \plus{} 1)(1 \plus{} a)} \plus{} \frac {c}{(a \plus{} 1)(1 \plus{} b)} \geq \frac {9}{16}$

1) $ \sum ab(a\plus{}b) \equal{} (a\plus{}b)(b\plus{}c)(c\plus{}a)\minus{}2abc \geq \frac 34 (a\plus{}b)(b\plus{}c)(c\plus{}a)$
equivalent to $ (a\plus{}b)(b\plus{}c)(c\plus{}a) \ge 8abc$ by am-gm

2) is also easy and equivalent to $ \frac {41}{3} \plus{} \frac 13 \ge 41(ab\plus{}bc\plus{}ca) \plus{} 9abc$

[quote=y-a-s-s-i-n-e]let $ a$ , $ b$ , $ c$ $ \geq 0$ , prove that :
1)
$ \frac {ab}{(a \plus{} c)(c \plus{} b)} \plus{} \frac {bc}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {ca}{(c \plus{} b)(c \plus{} a)} \geq \frac {3}{4}$

[hide=*]2) if $ a \plus{} b \plus{} c \equal{} 1$ :
$ \frac {a}{(b \plus{} 1)(1 \plus{} c)} \plus{} \frac {b}{(c \plus{} 1)(1 \plus{} a)} \plus{} \frac {c}{(a \plus{} 1)(1 \plus{} b)} \geq \frac {9}{16}$[/hide][/quote]$$\iff (a+b)(b+c)(c+a)\geq 8abc$$


Let $ ABCD$ be a trapezoid with $ BC\parallel AD$, and let $ O$ be the point of intersection of its diagonals $ AC$ and $ BD$. Prove that $ \left\vert ABCD\right\vert \equal{}\left(  \sqrt{\left\vert BOC\right\vert }\plus{}\sqrt{\left\vert DOA\right\vert }\right)  ^{2}$.

[hide="Source of the problem"][i]Source of the problem:[/i] exercise 8 in: V. Alekseev, V. Galkin, V. Panferov, V. Tarasov, [i]Zadachi o trapezijah[/i], Kvant 6/2000, pages 37-4.[/hide]

Very nice problem.

The equality is equivalent to $ (AOB) + (COD) = 2\sqrt {(BOC)(AOD)}$. Using $ \sin\angle BOC = \sin\angle AOB$ we can rewrite this as
\begin{eqnarray*} & & AO\cdot BO + CO\cdot DO = 2\sqrt {(BO\cdot CO)(AO\cdot DO)} \\
& \Leftrightarrow & (\sqrt {AO\cdot BO} - \sqrt {CO\cdot DO})^2 = 0 \\
&\Leftrightarrow & \frac {AO}{DO} = \frac {BO}{CO}\end{eqnarray*}
Which is true as $ \angle OBC = \angle ODA \wedge \angle OCB = \angle OAD\Rightarrow \triangle BOC\sim\triangle DOA$.

(EDIT) Remark: From this problem we can derive the following: in any trapezoid $ ABCD$ with $ BC\parallel AD$ we have $ (BOC)+(AOD)\ge (AOB)+(COD)$.

If call $ \frac{OC}{OA}\equal{}\frac{OB}{OD}\equal{}k$ then:
$ S_{AOB}\equal{}S_{COD}\equal{}k.S_{AOD},S_{BOC}\equal{}k^2.S_{AOD},S_{ABCD}\equal{}(k\plus{}1)^2.S_{AOD}$
Then it's OK!

Probably the problem is very famous. Here is an problem from Japan.

[color=blue]1961　Keio University entrance exam/Engineering[/color]

Let $ O$ be the point of intersection of its diagonals of a convex quadrilateral $ ABCD$ and let $ a^2,\ b^2\ (a > 0,\ b > 0)$ be the area of $ \triangle{AOD},\ \triangle{BOC}$ respectively. Denote the area of the quadrilateral by $ S$,

(1) Show that $ S\geq (a \plus{} b)^2$.
(2) Determine the shape of the quadrilateral when $ S \equal{} (a \plus{} b)^2$.

[quote="Darij Grinberg & Kunny"] [color=darkred]Let $ ABCD$ be a quadrilateral. Denote $ O\in AC\cap BD$.

Prove that $ \sqrt {[ABCD]} = \sqrt {[BOC]} + \sqrt {[DOA]}\Longleftrightarrow BC\parallel AD$.[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote $ [ABCD] = S$ and $ \{\begin{array}{ccc} [AOB] = u & , & [BOC] = x \\
\ [COD] = v & , & [DOA] = y\end{array}$.

Show easily that $ \boxed {\ xy = uv\ }$. Thus, $ \sqrt S = \sqrt x + \sqrt y$ $ \Longleftrightarrow$

$ x + y + u + v = x + y + 2\sqrt {xy}$ $ \Longleftrightarrow$

$ u + v = 2\sqrt {uv}$ $ \Longleftrightarrow$ $ u = v$ $ \Longleftrightarrow$ $ BC\parallel AD\ .$

[b]Remark.[/b] In any quad. $ ABCD$ exists the relation (near to evidence !) $ \sqrt S\ge\sqrt x + \sqrt y\ .$

[b]Remark.[/b] In the my opinion, this nice and easy old problem is for the middle-school students ![/color]

[hide="Proof."]
$ \frac ux = \frac {OA}{OC} = \frac yv\implies xy = uv\ .$[/hide]
[color=darkblue][b][u]A similar nice and easy old problem.[/u][/b] [/color]

[quote][color=darkred]Let $ ABCD$ be a quadrilateral and let $ d$ be a mobile line for which $ A\in d$

and $ \{\begin{array}{c}
X\in CB\cap d\ ,\ B\in (CX)\\\\
Y\in CD\cap d\ ,\ D\in (CY)\end{array}$. Prove that

$ [ABX]+[ADY]\ge [ABCD]$ with equality iff $ AX=AY\ .$[/color][/quote]
[color=darkblue][b][u]Here is a nice consequence of the initial proposed problem.[/u][/b][/color]

[quote="Virgil Nicula"][color=darkred] Let $ ABCD$ be a quadrilateral. Denote $ O\in AC\cap BD$. Prove that

$ 2\sqrt {[ABCD]} \ge \sqrt {[AOB]} + \sqrt {[BOC]} + \sqrt {[COD]} + \sqrt {[DOA]}$

with equality iff the quadrilateral $ ABCD$ is a parallelogram.[/color] [/quote]
[size=134][color=darkblue][b]I like very much these problems ![/b][/color][/size]