I have noticed that the AMC 10 seems to be favored over the AMC 12, so here is an analysis of AMC 10 vs. 12: (note that I am about to take out my analysis in the other topic)

Case A: You're an 11th or 12 grader, you can't take AMC 10, so this case is irrelevant.

Case B: You're in 7-10, and you got below the first index, but got $\ge$ the floor on the AIME. In this case, your qualification depends on how high your index is, regardless of whether you took the AMC 10 or AMC 12. However, because the AMC 10 is easier, you are more likely to have done better on the AMC and therefore gotten a higher index if you took the AMC 10. So here, the AMC 10 is favored over the 12.

Case C: You're in 7-10, and got an index above the first cutoff, but got an AIME score below everyone else in the first cutoff. If you took the 10, you would not get into the first cutoff seeing as you took the 10. If you took the 12, you would be in the first cutoff, qualifying for USAMO, and you would set the floor value. In this rare case, 12 would be better than 10.

Case D: You're in 7-10, made the floor (ie was not the sole person who set it), and got an index for the first cutoff. Here, if you had taken the 12, you would have made the first cutoff. If you had taken the 10, you would have had the floor or higher, and your index would have been fine, so you'd qualify. 10 vs. 12 doesn't matter in this case.

Case E: You're in 7-10, got an AIME score below the floor, and got an index below the first cutoff. Here you have no chance, so 10 vs. 12 doesn't matter.

I think we can conclude that taking the AMC 10 is favorable (for USAMO qualification) over the AMC 12?

Finally, I would like to give a quote:

[quote="AMC Website"]In advising students who desire to be selected for the USAMO whether to take the AMC 12 level contests or the AMC 10 level contests, it will be to their advantage to take the AMC 12 level contests.[/quote]

I strongly agree that the 10 is favored.  Yet, the website says........

[quote="Phelpedo"]I think the point is that if you're in 7-10th grade, taking the 12 gives you a chance to set the floor; otherwise the index isn't really relevant because it usually doesn't matter for 7-10th graders. I think this year was rare in that there was also a cutoff index for 7-10th graders who got an 8, because there were so many people with 8's. Keep in mind though that without the expansion, those people would not have made it anyway.[/quote]

The chance to set the floor is almost none existant and the fact is: [b]the expansion does exist,[/b], and the people who took the 10 had a large advantage over people like me :( .

Moral of story: always do things the easy way, take the easier test

[quote="absolutetruth"][quote="calc rulz"]I have noticed that the AMC 10 seems to be favored over the AMC 12, so here is an analysis of AMC 10 vs. 12.
[...]
I think we can conclude that taking the AMC 10 is favorable (for USAMO qualification) over the AMC 12?

Finally, I would like to give a quote:

[quote="AMC Website"]In advising students who desire to be selected for the USAMO whether to take the AMC 12 level contests or the AMC 10 level contests, it will be to their advantage to take the AMC 12 level contests.[/quote][/quote]

I could not agree more.  It definitely favors the AMC 10 over the AMC 12.  Why does the AMC website put out a misleading/wrong message??[/quote]

To screw up people like me? :?  :(

[quote="Phelpedo"]You don't necessarily have to set the floor, but taking the 12 enables you to qualify by being in the initial group. Look at nat mc, he got 7 on the AIME but was very close to setting the floor (113 index). Just a few more points on the 12 and he would have gotten in. And since the expansion is new this year, that line may have been applicable to years past, where there was no expansion.[/quote]

Well, if they can have seperate indices to adjust for the difficulty of the aimes, then obviously they can adjust between the 12 and the 10. :mad:

[quote="pieterminate"][/quote][quote="absolutetruth"][quote="calc rulz"]I have noticed that the AMC 10 seems to be favored over the AMC 12, so here is an analysis of AMC 10 vs. 12.
[...]
I think we can conclude that taking the AMC 10 is favorable (for USAMO qualification) over the AMC 12?

Finally, I would like to give a quote:

[quote="AMC Website"]In advising students who desire to be selected for the USAMO whether to take the AMC 12 level contests or the AMC 10 level contests, it will be to their advantage to take the AMC 12 level contests.[/quote][/quote]

I could not agree more.  It definitely favors the AMC 10 over the AMC 12.  Why does the AMC website put out a misleading/wrong message??[/quote][quote="pieterminate"]

To screw up people like me? :?  :([/quote]

I think they just simply didn't anticipate this happening  ;)

Yeah it was their first year dealing with th index

[quote="Phelpedo"][quote="pieterminate"][quote="solafidefarms"]Ya know, there's nothing preventing you from taking both, if you're that nervous about making it. Only a little money.[/quote]

REALLY!!??  How 'bout a bad school which only offers 1? :mad:[/quote]
You can always take it through another school, or yourself (I did this). It's a little extra work, but it's perfectly possible.

[quote="pieterminate"]Well, if they can have seperate indices to adjust for the difficulty of the aimes, then obviously they can adjust between the 12 and the 10.[/quote]
Er, well the difference here is that 10 and 12 are supposed to be different, unlike the AIMEs.[/quote]

Yep of course, travel a few hundred miles looking for an non-existant school from middle of nowhere.
And that's the point, they are being COUNTED the same.

[quote="randomdragoon"][quote="h_s_potter2002"]I thought the whole point of two AMC's was to give those who couldn't make it on the first date another chance to qualify.[/quote]
Or to give a chance to people who have no idea how the scoring system works to take both tests.[/quote]

Or.....give a chance only to the people in rich large schools to take two tests, while misleading the others... :?

[quote="h_s_potter2002"]I thought the whole point of two AMC's was to give those who couldn't make it on the first date another chance to qualify.[/quote]

This is off-topic. We're discussing the AMC 10 vs. AMC 12, not the "A" contest vs. the "B" contest.

[quote="towersfreak2006"]In qualifying for the AIME, something all USAMO hopefuls must nevertheless do, the AMC 12 is easier, statistcally.

AMC 10A had 1680 AIME qualifiers out of 61743 or rougly 2.7%.
AMC 10B had 870 AIME qualifiers out of 38420 or rougly 2.3%

AMC 12A had 17118 AIME qualifiers out of 76439 or roughly 22.4%.
AMC 12B had 10654 AIME qualifiers out of 50147 or roughly 21.2%

The statement given by the AMC website is not entirely untrue.[/quote]

Don't forget that one only must get a 100 on the AMC 12 to qualify for the AIME, while one must get a 120 on the AMC 10 to qualify for the AIME.

[quote="MathChauffeur"][quote="calc rulz"] ... I think we can conclude that taking the AMC 10 is favorable (for USAMO qualification) over the AMC 12? ... [/quote]As far as I can tell, AMC10 has nothing on AMC12:

2. For 10th grade and below, only floor value matters. Prof Dunbar mentioned 212 being the lowest index among the floor qualifiers regardless of 10/12 -- for info only, it had nothing to do with qualifying.
[/quote]

The 212 was a "tiebreaker" between those above the floor.  So for example, I had 126.5 on amc 12 and 8 on aime, but didn't make the 212 index~~~~> no qualify....

 And remember: the 10 is easier than the 12~~~>easier to achieve index of 212.....

[quote="MathChauffeur"][quote="calc rulz"] ... I think we can conclude that taking the AMC 10 is favorable (for USAMO qualification) over the AMC 12? ... [/quote]As far as I can tell, AMC10 has nothing on AMC12:

1. AMC12 is considered in the first cut-off, AMC10 is not.

2. For 10th grade and below, only floor value matters. Prof Dunbar mentioned 212 being the lowest index among the floor qualifiers regardless of 10/12 -- for info only, it had nothing to do with qualifying.

3. AIME qualification by AMC12 is (at least should be) easier for those with USAMO aspiration.[/quote]

1. My point is that anyone who qualified by the first cutoff and who could have taken the AMC 10 would have still qualified by AMC 10, except in the rare case of the person who set the floor, which is irrelevant for the general case.

2. It did have to do with qualifying. I, a 9th grader, got an 8 on the AIME but did not qualify for USAMO -.-

3. People who are borderline for AIME don't tend to qualify for USAMO. As well, even if it's easier to qualify for AIME via AMC 12 (and I do agree that it is), USAMO contenders who took the AMC 10 still have a better shot at USAMO.

[quote="towersfreak2006"]I didn't. It's always fun to crunch some statistics yourself:
http://www.unl.edu/amc/e-exams/e5-amc10/e5-1-10archive/2006-10a/06-stats10.html
http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2006-12a/06-stats12.html

And what you say shows the AMC 12 as being more favorable.[/quote]

How so?

[quote="calc rulz"] ... 2. It did have to do with qualifying. I, a 9th grader, got an 8 on the AIME but did not qualify for USAMO -.- ... [/quote]Wow, my apology. I think I misunderstood Prof. Dunbar's statement.

To get a 101, which is qualifying via the AMC 12, one needs to at least answer 11 correctly and skip 14. To get a 122, which is qualifying via the AMC 10, one needs to at least answer 17 correctly and skip 8.

Now, considering that 9 of the first 11 questions on this year's AMC 12A and 10A were the same and that the two different problems were #7 and #9 on the 12A, which sounds easier to do? The statistics do not lie...

Of course, this is still talking about qualifying for the AIME, not the USAMO, but as I said earlier, one has to qualify for the former first before attempting to do so on the latter.

I do agree that the USAMO selection process favors the AMC-10 under the "212-system", but there is some truth in what the site said.

[quote="towersfreak2006"]
Of course, this is still talking about qualifying for the AIME, not the USAMO, but as I said earlier, one has to qualify for the former first before attempting to do so on the latter.
[/quote]

Yes, of course it's easier to qualify for the aime by the 12.........But someone who qualifies for the USAMO usually doesn't have any trouble qualifying for the AIME.


Word it better. I don't think it's answer-able right now ;)

[quote="236factorial"]on a coordinate plane, 4 circles or radius 4 meet in the origin, making 4 overlap spots. what is the total area of the overlap spots?

is this too vague without a diagram? source:mathcounts national competition, 2001, sprint round number 29.[/quote]
I need a diagram for sure.

I get what you mean.  Well there are 4 overlapping areas of the same area.  SPlit each into 2 parts.  This part is just a sector minus a triangle.  It is an isoseles so the angle is 90. One of these eight areas has area:
$\frac{4^2\pi}{4}-{4^2}{2}=4\pi-8$
There are 8.  Total area:
${\{32\pi-64}$

I'm assuming this is what you're saying. (see dia)

[hide]
Consider a semi-circle with radius 4 that has the same overlapping region.  The area of the half with the shaded region is $((4^2\pi)/2)/2$.  You can subtract the area of the right triangles defined by the radi which are tge 2 sides of the 1/2 of the semi-circle.  That leaves you with $4\pi-4*4/2 = 4\pi-8$.  Each of the overlapping regions have double that amount, which gives us $8\pi-16$.  There are 4 overlapping regions, so the area of the overlapping regions is:

$32\pi-64$

~ 36.530964914873383630804588264944
[/hide]

Edit: Thought you meant overlapping part was a semicircle

It said nothing about semicircles :roll: .


function
advanced fields
advanced fields unsolved

By using mobius inversion's formula how can I prove that

$ P(x) = \sum_k^\infty 1/kp(x^{1/k})$
where $ p(x) = \sum_k^\infty u(k)/kP(x^{1/k}),$ here u is the mobius function.

sorry it should be

 By using mobius inversion's formula how can I prove that

P(x) = sum_k 1/kp(x^{1/k})
where p(x) = sum_k u(k)/kP(x^{1/k}), here u is the mobius function.

here both sums are infinte.


Aki el problema 5:

Sea ABC un triangulo acutangulo, y AD, BE, CF sus alturas. La circunferencia con diametro AD corta a los lados ABy AC en M y N respectivamente. Sean P y Q los puntos de interseccion de AD con EF y MN respectivamente. Demuestra que Q es el punto medio de PD.

 :lol: [/img]

Sea $R$ el punto de interseccion de $FC$ y $MN$, $T$ la intereccion de $BE$ y $MN$, con el cuadrilatero inscriptible $RDCN$ tenemos $\measuredangle DRC = 90^\circ$, en consecuencia $MDRF$ es un rectángulo, de lo cual: $MD=FH+HR$

$DMN$, $HRT$, $HFE$ son semejantes: $\frac{FH}{PH}=\frac{HR}{HQ}=\frac{MD}{QD}$

Luego: $PH+HQ=QD$ $\Rightarrow$ $\boxed{PQ=QD}$

Nota: $H$ es el ortocentro de $ABC$

Se puede plantear un resultado más general:

Sea $ X$ un punto del plano de $ \triangle ABC$ y $ \triangle DEF$ su triángulo ceviano. $ D \in BC, $ $E \in CA$  y $F \in AB.$ Las paralelas por $ D$ a $ BE$ y $ CF$ cortan a $ AC,AB$ en $ N,M.$ $ AD$ corta a $ EF$ y $ MN$ en $ P,Q.$ Entonces $ Q$ es punto medio de $ PD.$

Note que $ \triangle DMN \sim \triangle XFE$ son homotéticos con centro de homotecia $ A$ y coeficiente $ \frac {AD}{AX},$  por tanto $ \frac {QD}{PX} \equal{} \frac {AD}{AX}.$

Por teorema de Thales se tiene $ \frac {PQ}{AP} \equal{} \frac {EN}{AE} \equal{} \frac {XD}{AX}$

Combinado con la anterior resulta:

$ \frac {PQ}{QD} \equal{} \frac {AP}{PX} \cdot \frac {XD}{AD}.$ Pero como $ (A,X,P,D )=-1$ $  \Longrightarrow \ \frac {AP}{PX} \equal{} \frac {AD}{XD}$

Lo cual implica $ PQ \equal{} QD$ como se queria probar.


1) There are $17$ scientists who write to all $16$ of each other. In their letters, there are only $3$ topics to be discussed. But between $2$ scientists, they will only discuss $1$ topic  with each other. Prove that there are at least $3$ scientists,who discuss the same topic among them.

2) in a circle with radius $1$, inscribed a square and a triangle. If the 7 points are on different spots on the circumference of the circle, prove that the arc length between any of the 2 points are not more than $\frac{\pi}{12}$

first one should be classical

We pick a scientist and call it $A$ . Since $A$ wrote to other 16 , by pigeon hole , we can say that A wrote to 6 persons in the set $S\in (B,C,D,E,F,G)$ with question $I$ .

Now we consider $B$ . If $B$ wrote question $I$ to any others in $S$ , then we are done . Else by pigeon hole , there are exist $3$ of them in $S$ say $C,D,E$ whom $B$ wrote question $II$ to them .

Now if $C$ wrote question $I$ or $II$ to $D$ , we are done . Else say $C$ wrote question $III$ to $D$ . Similar argument we say that $D$ wrote question $III$ to $E$ . Now looking at $C$,$E$ . No matter what question $C$ wrote to $E$ , we will always have a cycle .

2 ) Not clear ... what is the use of the square and triangle thing ?

[quote="shyong"]2 ) Not clear ... what is the use of the square and triangle thing ?[/quote]

Inscribe a square and a triangle in a circle

[quote="JBL"]The statement of 2 is wrong.  It should be, "Inscribe a square and a triangle in a circle of radius 1.  Show that among the 7 vertices of the two polygons there are two which determine an arc of length at most $\frac\pi{12}$."[/quote]

sorry  :blush: , JBL is right. was in a hurry when i translated the question. so, what's the solution?


OK.  So I was doing MC practice on Thursday, we did the 2005 chapter sprint, and here was the question:

[b]A play has two male roles two female roles and two roles that can be assigned to either gender.  Only a man can be assigned a male role and only a woman can be assigned a female role.  If five men and 6 women audition, in how many ways can the 6 roles be assigned?[/b]

So I thought I could do 5C2(5 males two roles) times 6C2(6 females two roles) times 7C2(7 males/females remaining, 2 neutral roles) which equals 3,150.   which equals 3,150.  The answer key says I'm wrong.  What did I do wrong?

The answer's right here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=326057

Also, you forgot that they could be in distinct orders, so multiply each comboniation by $ 2!$.


Find all positive reals $x,y,z$ such that

\[ \left\{ \begin{array}{ll} x^3+y^3+z^3 = x+y+z\\ x^2+y^2+z^2=xyz\\ \end{array} \right. \]

I've been thinking also what happens if $x,y,z$ are just reals :?

[quote="mathmanman"]The (missing) system should be :
$\begin{cases}x^3 + y^3 + z^3 = x + y + z \\ x^2 + y^2 + z^2 = xyz \end{cases}$. ;)

And, it's 'easy' to see that it has no solution (hint : AM-GM), which leads us to Valentin's question where $x, y, z \in \mathbb{R}$. :)[/quote]
x=y=z=1?
I am sorry.

[quote="Rust"]x=y=z=1?
I am sorry.[/quote]

$1^2+1^2+1^2\neq 1\cdot 1\cdot 1$

$\begin{cases}x^3 + y^3 + z^3 = x + y + z \\ x^2 + y^2 + z^2 = xyz \end{cases}$
Let $x\ge y\ge z.$
From (1) $\Longrightarrow \ x\ge 1.$ From (2) $\Longrightarrow \ y,z\ge 1.$ From (1) contradition.


Please help. Where can I found formulas for different elements of the triangle !!! with proofs !!! (not basic ones, but like $ GI,IH,NI$ and others).

[b][color=brown]Thanks....[/color][/b]

Two of my favorite books for geometry of triangle are:
T.Lalescu: Geometria triunghiului, Ed. Apollo, Craiova 1993 and
D.Branzei: Competentza si performantza in geometrie (Vols. 1 and 2), Ed. MINIED, Iasi
Another complete set of formulae with distances between different points of the triangle was published in Gazeta Matematica by Viorel Bandila (Gazeta Matematica nr.2, 1985)


Let $ P(x)\equal{}x^n\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\dots\plus{}a_1x\plus{}1$-be a polynomial with nonnegative integer coefficients,such that $ a_k\equal{}a_{n\minus{}k}$,for all
$ 1\leq k\leq n\minus{}1$.Prove that there exist infinitely many pairs of natural numbers $ x$ and $ y$,such that
$ y|P(x)$ and $ x|P(y)$.

[hide="Hint"]If $ (x_k,y_k)$ is a pair satisfy that condition then we can find a pair $ (x_{k\plus{}1},y_{k\plus{}1})$ also satisfy. What is $ (x_{k\plus{}1},y_{k\plus{}1})$?[/hide]

[quote="N.T.TUAN"][hide="Hint"]If $ (x_k,y_k)$ is a pair satisfy that condition then we can find a pair $ (x_{k \plus{} 1},y_{k \plus{} 1})$ also satisfy. What is $ (x_{k \plus{} 1},y_{k \plus{} 1})$?[/hide][/quote]
Could you please post complete solution?
Thank you.

[quote="N.T.TUAN"][hide="Hint"]If $ (x_k,y_k)$ is a pair satisfy that condition then we can find a pair $ (x_{k \plus{} 1},y_{k \plus{} 1})$ also satisfy. What is $ (x_{k \plus{} 1},y_{k \plus{} 1})$?[/hide][/quote]
I'm waiting for your solution,dear N.T.TUAN,so could you please post your solution?


When all possible diagonals are drawn from one vertex of a 50-sided convex polygon, how may triangles are formed?

SHOW UR WORK

p.s. this is from 1995-1996 CHAPTER COMPETITION TEAM ROUND

dude look in miller mathcounts i've posted a similar question.

Not sure, but it looks as if the answer is simply C(50, 3), since you are choosing any 3 vertices from the 50 available.

I get [hide]48[/hide].

[hide]You are only drawing diagonals from one point, so you will draw 50-3 or 47 diagonals.  Each diagonal adds one triangle, with the final one adding two.[/hide]

fiery: if you were to do this with a square, or a pentagon, would you still get c(4,3) and c(5,3) respectively?

my bad, bookworm is correct. I didn't see the "one" in "drawn from one vertex", heh.

[quote="Monkeyanator1"]Ya, bookworm's response seems correct to me. I've done this MathCounts problem before, and that was the answer.[/quote]
Welcome to the forums Monkeyanator1!

For future reference, please do not "revive" posts,(comment on posts that are very old. This was from 2003) unless maybe they aren't solved. In this case, the problem has been solved sufficiently.

Oh, I'm extremely sorry. I'm [i]extremely[/i] new to this site. I was just looking for some problems, so I just looked at the older posts. Sorry again. Oh, by the way, please join my probability marathon!


inequalities
Divisors
number theory
IMO
IMO 2002
IMO Shortlist

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[quote="pavel25"]$d_k = {n\over d_1}$
$d_{k-1} = {n\over d_2}$
$d_1 = {n\over d_k}$
So we can write it like this:
$d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}$
The Maximum that can be is:
${{n^2}\over 1*2}+{{n^2}\over 2*3}+\,\ldots\,+{{n^2}\over (n-1)n}$
We can see that the sum starting with the form:
${m\over {m+1}}{n^2}$
We check what happens if we add the next term:
${m\over {m+1}}{n^2} + {1\over {(m+1)(m+2)}}{n^2} = {{(m+1)^2}\over {(m+1)(m+2)}}{n^2} = {{m+1}\over {m+2}}{n^2}$
From here is easy to define that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}<{n^2}$

$d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}$ is a divisor of ${n^2}$ when $n$ is a prime number:

$d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k} = n$

if $n$ isn't a prime we have at least 3 divisors of $n$
We proved that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}<{n^2}$

So if it's a divisor of ${n^2}$ the maximum that can be: ${{n^2}\over d_2}$

$d_1 = 1$

$d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k} = {{n^2}\over d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k} > {{n^2}\over d_2}$
We see from here that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ is a divisor of ${n^2}$ only if $n$ is a prime![/quote]

I have asked this question as another link sorry for the same question <http://www.mathlinks.ro/Forum/viewtopic.php?p=531646#531646>

:-) see i was right about the imo 2002

davron

[hide="Sorry to revive"]
Notice that \[ S\equal{}\sum d_id_{i\plus{}1}<n^2\iff \sum\frac{1}{d_id_{i\plus{}1}}<1\]
Luckily, \[ \sum\frac{1}{d_id_{i\plus{}1}}\leq\sum\frac{d_{i\plus{}1}\minus{}d_i}{d_{i\plus{}1}d_i}\equal{}\sum\left(\frac{1}{d_i}\minus{}\frac{1}{d_{i\plus{}1}}\right)\equal{}1\minus{}\frac{1}{n}<1\]
Thus, the first part is established.

Now, suppose that $ S|n^2$. Then, $ 1<n^2/S\leq \frac{n^2}{d_{k\minus{}1}d_k}\equal{}d_2$ is a divisor of $ n^2$ also. Since the smallest non trivial divisor of $ n^2$ is the same as the smallest non trivial divisor of $ n$ which is $ d_2$, it follows that $ n^2/S\equal{}d_2$. Hence, $ S\equal{}d_{k\minus{}1}d_k$ and therefore $ k\minus{}1\equal{}1$, establishing that $ n$ is prime.
To establish the converse, suppose that $ n$ is prime. Then clearly the result holds, establishing that the second condition is equivalent to $ n$ being prime.[/hide]

Notice that because of the famous Cauchy Inequality applied to $d_1,...,d_{k-1}$ and $d_2,...,d_k$ we will have

$d_1d_2+...+d_{k-1}d_k \le \left( d_1^2+...+d_{k-1}^2 \right)^2 \left( d_2^2+...+d_k^2 \right)^2$.

Let $X=d_1^2+...+d_k^2$, we will have that

$d_1d_2+...+d_{k-1}d_k \le \left( X-n^2 \right)^2 \left( X-1 \right)^2$

So, in order to finish the first part it is enough to prove

$X^2-(n^2+1)X+(n^2-n^4) < 0$

which simplifies, by the general equation, to proving that

$X < \displaystyle\frac{n^2+1+\sqrt{5n^4-4n^2+1}}{2}$

Since $\sqrt{5n^4-4n^2+1} > (2n^2-1)$, it is enough to prove that

$X \le \displaystyle\frac{3n^2}{2}$

Let $n=p_1^{e_1}...p_m^{e_m}$, we will have

$X=\prod_{i=1}^m (1+p_i^2+...+p_i^{2e_i})$

and so it will be enough to prove that, for each $i \le m$, if $p=p_i$ and $e=e_i$,

$3p^{2e}/2 \ge 1+p^2+...+p^{2e}$

This is equivalent to

$p^{2e}/2 \ge 1+p^2+...+p^{2e-2} = \displaystyle\frac{p^{2e}-1}{p^2-1}$

Notice that $p^2-1 \ge 4-1 = 3$ and so the above is true.

So we are done with the first part.

Now for the second part, assume that $X=d_1d_2+...+d_{k-1}d_k | n^2$. If $n$ is not prime then $k >2$ and so $X > d_kd_{k-1}=n^2/p$, where $p$ is the smallest prime divisor of $n$ (which exists because $n>1$). And so if $X | n^2$ then there exists a $q$ such that $qX=n^2$. Notice that $X < n^2$ by the first part and so $q > 1$. Since $q | n$, then $q \ge p$ and we get $pX \le n^2$ which is false because $X > n^2/p$.

So $n$ must be prime, and it is easy to see $X=1*n|n^2$.

Part a: Notice that $d_i d_j = n $ if and only if $i+j = k+1$
So we have that $d_1d_2 +d_2d_3+...+d_{k-1}d_k < n^2 \iff$
$ d_{k-1}d_k (d_1d_2+...+d_{k-1}d_k) < n^2 (d_{k-1}d_k)$
Wich is equal to
$n^2 + d_{k-1}d_k(d_2d_3+...+d_{k-1}d_k) < RHS$
Multiplying by the other terms we have that
$ n^2(d_1d_2)(...)(d_{k-2}d_{k-1}) + n^2 (d_2d_3+...+d_{k-1}d_k) < n^2 (d_1d_2)(...)(d_{k-1}d_k)$
Wich is equal to
$n^2(( \frac {(k-1) n^2} {d_{k-1}d_k})+(d_2d_3+...+d_{k-1}d_k)) < n^2 ((k-1) n^2))$
we have that clearly $d_id_{i+1}<n^2$ for every i
So we just have to show that
$\frac {(k-1) n^2} {d_{k-1}d_k} \le n^2 \iff $
$(k-1) \le d_{k-1}d_k$
We know that $ 1< d(n) < n$ so clearly this inequality holds
( $k-1<k=d(n)<n\le d_{k-1}n$)

For part b: we have that the greatest divisor of $n^2$ is $\frac {n^2} {d_2} = d_{k-1} d_k $
And by part a we know that $n^2 > LHS \ge d_{k-1} d_k $ and equality holds only when n is a prime so thats the only posible case

(a) Clearly, $d_id_{k+1-i}=n$, so
\[ d_id_{i+1} = \frac{n^2}{d_{k+1-i}d_{k-i}} = \frac{n^2}{d_{k+1-i}-d_{k-i}}\left(\frac{1}{d_{k-i}} - \frac{1}{d_{k+1-i}}\right) \le n^2\left(\frac{1}{d_{k-i}} - \frac{1}{d_{k+1-i}}\right), \]
since $d_{k+1-i} - d_{k-i}\ge 1$. Summing, we get
\[ \sum_{i=1}^n d_id_{i+1} \le n^2 \sum_{i=1}^n \frac{1}{d_{k-i}} - \frac{1}{d_{k+1-i}} = n^2\left(\frac{1}{d_1}-\frac{1}{d_k}\right)=n^2(1-1/n) < n^2.\]
(b) Let $S=d_1d_2+\cdots+d_{k-1}d_k$. First consider the case where $n$ is composite. We know $d_2$ is the smallest prime divisor of $n$. Let $d_2=p$. Since $d_{k-1}d_k=\tfrac{n}{p}\cdot n = \tfrac{n^2}{p}$, so $S>n^2/p$. But $p$ is also the smallest prime divisor of $n^2$, so $n^2/p$ is the largest proper divisor of $n^2$. Therefore, $S=n^2$, which is impossible, since it is at most $n^2(1-1/n)$. Hence, no composite $n$ works. Now consider the case where $n$ is prime. Then $d_1=1,d_2=n$, so $S=d_1d_2=n$, which divides $n^2$. Therefore, $S\mid n^2$ if and only if $n$ is prime.

We always have \begin{align*} 	d_k d_{k-1} + d_{k-1} d_{k-2} + \dots + d_2 d_1 	&< n \cdot \frac n2 + \frac n2 \cdot \frac n3 + \dots \\ 	&= \left( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots \right) n^2 = n^2. \end{align*} This proves the first part.

For the second, we claim that this only happens when $n$ is prime (in which case we get $d_1 d_2 = n$). Now assume $n$ is not prime (meaning $k \ge 2$) and let $p$ be the smallest prime dividing $n$. Then \[ d_k d_{k-1} + d_{k-1} d_{k-2} + \dots + d_2 d_1 	> d_k d_{k-1} = \frac{n^2}{p} \] exceeds the largest proper divisor of $n^2$, but is less than $n^2$, so does not divide $n^2$.

[quote=orl]Let $n\geq2$ be a positive integer, with divisors $1=d_1<d_2<\,\ldots<d_k=n$.  Prove that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ is always less than $n^2$, and determine when it is a divisor of $n^2$.[/quote]

Notice that \begin{align*} \sum^{k-1}_{i=1} d_id_{i+1} &=n^2 \cdot \left( \sum^{k-1}_{i=1} \frac{1}{d_id_{i+1}}\right) \\ & \le n^2 \cdot \left( \sum^{k-1}_{i=1} \frac{(d_{i+1}-d_i)}{d_id_{i+1}} \right) \\ &= n^2 \cdot \left(\sum^{k-1}_{i=1} \left(\frac{1}{d_i}-\frac{1}{d_{i+1}} \right) \right) \\ &= n^2 \cdot \left(1-\frac{1}{n}\right) <n^2, \end{align*} so the first assertion is true. Observe that $d_{k-1}=\frac{n}{p}$ where $p$ is the smallest prime factor of $n$; so if $k>2$ then $$1<\frac{n^2}{\sum^{k-1}_{i=1} d_id_{i+1}}<\frac{n^2}{d_{k-1}d_k}=p,$$ contradicting that this quotient is a divisor of $n^2$. So, $k=2$ and $n=p$. For all prime numbers $n$, the assertion holds; hence these are the only solutions.

Note that
\[S:=d_1d_2+\cdots+d_{k-1}d_k=n^2\left(\frac{1}{d_1d_2}+\cdots+\frac{1}{d_{k-1}d_k}\right)\le n^2\left(\left(\frac{1}{d_1}-\frac{1}{d_2}\right)+\cdots+\left(\frac{1}{d_{k-1}}-\frac{1}{d_k}\right)\right),\]
so
\[S\le n^2(1-1/n)=n^2-n<n^2,\]
as desired.

Now suppose that $S\mid n^2$. Any divisor of $n^2$ can be written as a product of two divisors of $n$, so suppose $S=d_ad_b$ where $b>a$. We'll show that this isn't possible for size reasons. To do so, we have the following ``smoothing'' lemma.

[u]Lemma:[/u] If $0<x\le z\le y$, then $xy<xz+zy$.

[u]Proof:[/u] Over the interval $[x,y]$, the extrema of the linear function $z(x+y)$ will be at the edges, so
\[x^2+xy\le xz+zy\le y^2+xy,\]
so certainly $xy<xz+zy$. $\blacksquare$

Suppose for now that $b-a\le 2$ Using the lemma, we see that
\[d_ad_b<d_ad_{a+1}+d_{a+1}d_{a+2}+\cdots+d_{b-1}d_b,\]
so $d_ad_b<S$. If $b-a=1$, then the only way $S=d_ad_{a+1}$ is if $n$ has only two factors, so $n$ is prime. Primes clearly work as $S=1\cdot p\mid p^2$. Thus, the answer is $\boxed{n\text{ is prime}}$.

Cute problem :)
Notice that $d_k \leq \frac n1, d_{k-1} \leq \frac{n}2, d_{k-2} \leq \frac{n}3, $ and so on.
Thus the sum $X = d_kd_{k-1}+d_{k-1}d_{k-2}+...+d_2d_1 \leq \frac{n^2}{1 \cdot 2} + \frac{n^2}{2 \cdot 3} +...+ \text{(k-1 terms)} < n^2(\frac 1{1 \cdot 2} +\frac 1{2 \cdot 3}+... \text{(infinite sum)})=n^2( \frac 11 -\frac 12 + \frac 12 - \frac 13 +\frac 13 -... )=n^2$.
$=>$ $a$ part over.
After a little playing around, or even otherwise, we see that the only $n$ that'll work should be prime $n$- once we realise this, it basically becomes a one-liner: consider $n$ non-prime with smallest divisor $p$. Notice that the largest divisor of $n^2<n^2$ must be $\frac{n^2}p$, and as our sum $X<n^2$ but $d_k d_{k-1} = \frac{n^2}p =>$ if there are more than $2$ factors, $X> \frac{n^2}p$ can't divide $n^2$. Verifying, we see that each $n$ prime indeed works, hence the only $n$ that work are primes.

We have $d_k\le n$, $d_{k-1}\le \frac{n}{2}$, and continuing, $d_i\le \frac{n}{k+1-i}$. Then telescoping sums proves the first part. For the second part, let $p$ be the smallest prime dividing $n$, and note that the sum must be $\frac{n^2}{i}$ for some $1\le i<p$ since it is greater than $\frac{n^2}{p}$. Then we are done since this is never an integer. $\blacksquare$

(unless $n$ is prime oops)

For the first part, we have that $$d_kd_{k-1} + d_{k-1}d_{k-2} + \cdots + d_2 d_1 \le \frac{n}{1} \cdot \frac{n}{2} + \frac{n}{2} \cdot \frac{n}{3} + \cdots + \frac{n}{n-1} \cdot \frac{n}{n}  = n^2-n < n^2.$$
For the second part, let $p$ be the smallest prime divisor. We have that the sum must be less than or equal to $\frac{n^2}{p}$. We have that $$d_1d_2 + \cdots + d_{k-1}d_{k} \ge d_{k-1}d_k = \frac{n^2}{p},$$ with equality iff $n$ has two divisors or $n$ is prime. This works so we get the answer is $n$ prime only.

$d_{k-1}\leq \frac{d_k}{2}, d_{k-2}\leq\frac{d_k}{3},...,d_1\leq \frac{d_k}{k}$ so $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k\leq \frac{d_k^2}{k}\cdot\frac{d_k^2}{k-1}+\frac{d_k^2}{k-1}\cdot\frac{d_k^2}{k-2}+...+\frac{d_k^2}{2}\cdot\frac{d_k^2}{1}=d_k^2(1-\frac{1}{k})<d_k^2$
For the second part, let $p$ be the smallest prime divisor of $n$, and let $m=d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k\mid d_k^2$.
Since $1<\frac{d_k^2}{m}\leq p$, we must have $m=\frac{d_k^2}{p}=d_{k-1}d_k$, which means $d_{k}=p$. Therefore, it only works when $n$ is prime.

Let $f(n)$ denote $d_1d_2+d_2d_3\cdots +d_{k-1}d_k.$ Clearly, we have $$f(n)< n(n/2)+(n/2)(n/3)+(n/3)(n/4)\cdots =n^2(1/2+1/6+1/12\cdots)=n^2. $$ (inequality is strict since of course $n$ is finite).\\

We claim that $f(n)$ divides $n^2$ if and only if $n$ is prime. If $n$ is prime, $f(n)=n.$\\

Let $p$ be the smallest prime that divides $n$. Note that if $f(n)<n^2$, then in order for $f(n)$ to divide $n^2$, we must have $$f(n)\leq \frac{n^2}{p}.$$ However, the product of the two largest factors, $n$ and $n/p$, already contribute $n^2/p$, so there cannot be any factors other than those two, so we are done.

For the first part, note that
\[
\sum_{i = 1}^{k-1}\frac{d_i}{n}\cdot \frac{d_{i+1}}{n} = \sum_{i = 1}^{k-1}\frac{1}{d_id_{i+1}}< \sum_{i =1}^{k-1}\frac{1}{i(i+1)}< \sum_{i =1}^\infty\frac{1}{i(i+1)} = 1.
\]
Now, for the second part, we claim that the divisibility holds if and only if $n$ is prime. It is easy to see that all such $n$ work, so we now show that others fail. Indeed, let $p$ be the smallest prime divisor of $n$. Then, we have
\[ \sum_{i = 1}^{k-1}d_id_{i+1}\ge d_{k-1}d_k = \frac{n}{p}\cdot n = \frac{n^2}{p},\]
which is the largest proper divisor of $n^2$. We have already shown that the sum is less than $n^2$, so for it to divide $n^2$, it must be exactly equal to $n^2/p$. This means equality holds in the inequality above, so $d_{k-1}$ and $d_k$ must be the only divisors of $n$. Hence, $\tau(n) = 2$, which means $n$ is necessarily prime.

Part 1: Note that $d_1d_2$ is the greatest term, so to maximize the sum, we need $d_2 = d_1/2$.  Anything greater than $d_1/2$ would not be a divisor of $d_1$(not including itself) since division would result in a result between 1 and 2 which is not an integer.  Similarly, the greatest value possible for $d_3$ is $d_1/3$ as it cannot equal $d_2$.  Thus, the greatest possible sum would be $(d_1)(d_1/2) + (d_1/2)(d_1/3) + (d_1/3)(d_1/4)+...+(d_1/(d_1-1))(d_1/d_1) < (d_1)^2$ .  Because this is the optimal case, the sum is always less than $n^2$.

Part 2:  Let $p$ be the least prime divisor of $n$.  Note that $d_1d_2 = (n)(n/p)$ as $n/p$ is the greatest divisor of $n$ that is less than itself.  In order for the sum to divide $n^2$, we need $\frac{n^2}{(n)(n/p)+...}$ to be an integer.  The maximum value of $\frac{n^2}{(n)(n/p)+...}$ is p and the equality is when $n=p$ as it would result in just two factors.  Also, $\frac{n^2}{(n)(n/p)+...}$ is greater than 1 as we proved in the first part.  Note that if $n$ is composite, then an integer less than $p$ would divide $n^2$ contradicts the statement that $p$ is the smallest prime that divides $n$.  Therefore, $n$ is prime and we can show that this condition holds for primes as $d_1=n$ and $d_2=1$, so $d_1d_2 = n|n^2$.

Observe that $d_i = n/d_{k-i+1}$, so that $d_1d_2 + d_2d_3 + \cdots + d_{k-1}d_k = n^2 \left(\sum_{i=1}^{k - 1} \frac{1}{d_id_{i+1}} \right)$. Hence it suffices to show that the inner RHS sum is less than $1$.
\begin{align*}
d_1d_2 + d_2d_3 + \cdots + d_{k-1}d_k &= \sum_{i=1}^{k - 1} \frac{1}{d_id_{i+1}} \\
& \leq \sum_{i=1}^{k - 1} \left(\frac{1}{d_i} - \frac{1}{d_{i + 1}} \right) \\
&= 1 - \frac{1}{d_k} < 1.
\end{align*}

For the second part of the problem, I claim that only $\boxed{n = p}$, for all primes $p$ work. Now from the above, we find that this sum $S$ is strictly less than $1$. Therefore in order to have $n^2S < n^2$ and to divide $S$, we have that $S$ is at most $\frac{n^2}{d_2}$. Therefore \[n^2S > \sum_{i=1}^{k - 1} \frac{n^2}{d_id_{i+1}} \geq \frac{n^2}{d}, \] so that $\boxed{k = 2}$ if we want $n^2S \mid n^2$. Hence $n$ has two divisors being $1$ and itself, hence prime. $\blacksquare$

we can easily see that all $d_x=\frac{n}{d_{k+1-x}}$
then, $d_1d_2+d_2d_3+\cdots+d_{k-1}d_k$ can be expressed as $\frac{n^2}{d_kd_{k-1}}+\frac{n^2}{d_{k-1}d_{k-2}}+\cdots+\frac{n^2}{d_1d_2}$
it is clear that the maximum for this sum is $\frac{n^2}{1*2}+\frac{n^2}{2*3}+\cdots+\frac{n^2}{(n-1)(n)}$
this can be expressed as $n^2(\frac{1}{1*2}+\frac{1}{2*3}+\cdots+\frac{1}{(n-1)(n)}$
which is $n^2(1-\frac{1}{2}+\frac{1}{2}-\frac{2}{3}+\cdots+\frac{1}{n-1}-\frac{1}{n})$
which is equal to $n^2$
(second part of this proof tba)

I have extreme amounts of hating math - not because of this problem but because gradually math is more boring and i want to deviate to do CS or physics.

$\frac{n^2}{d_kd_{k-1}}+\frac{n^2}{d_{k-1}d_{k-2}}+...+\frac{n^2}{d_1d_2}\le n^2(\frac1{1\cdot 2}+\frac1{2\cdot 3}+...)=n^2(1-\frac1n)<n^2$ UNLESS n is prime, then the LHS=n, which works, and it's the only solution since if the smallest prime p dividing n is not n then to divide n we  must have $LHS\le\frac{n^2}p$, contradiction since n and n/p already multiply to n^2/p in the sum, and we at least have the other factor 1 with the second smallest factor.

Notice that $$d_1d_2+d_2d_3+\cdots+d_{k-1}d_k = n^2\left(\frac 1{d_1d_2} + \frac 1{d_2d_3}+\cdots + \frac 1{d_{k-1}d_k}\right).$$ On the other hand, as $\frac 1{d_1d_2} < \frac 1{d_1} - \frac 1{d_2}$, this is strictly less than $n^2$.

On the other hand, as $d_{k-1}d_k = \frac{n^2}p$ where $p$ is the smallest prime divisor of $n$, it follows that the sum must equal exactly $\frac{n^2}p$. This occurs if and only if $n$ is prime.

Note that $d_k\le \frac n1$, $d_{k-1} \le \frac n2$, $d_{k-2} \le \frac n3$, $\ldots$, $d_2 \le \frac{n}{k-1}$, $d_1\le \frac{n}{k}$. In general, $d_i\le \frac{n}{k-i+1}$. Thus, we have the inequality$$d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k\le n^2\sum^{k-1}_{i=1}\frac 1{i(i+1)}.$$The summation telescopes into $1-\frac 1k$, so we have $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k\le n^2\left(1-\frac 1k\right)<n^2$, as desired.

Now, I claim that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ divides $n$ only if $n$ is prime. Suppose the smallest divisor of $n$ other than $1$ is $p$. Obviously, $p$ is prime. Note also that $p$ is the smallest divisor of $n^2$ other than $1$. Then, we have that $d_2 = p$ and $d_{k-1} = \frac np$. We then have that our sum is equal to$$p+\frac{n^2}{p} + d_2d_3+d_3d_4\ldots + d_{k-2}d_{k-1}>\frac{n^2}{p}$$$$\implies \frac{n^2}{p+n^2+d_2d_3+d_3d_4+\ldots + d_{k-2}d_{k-1}}<p.$$However, if $p+\frac{n^2}{p} + d_2d_3 + d_3d_4+\ldots + d_{k-2}d_{k-1}\mid n^2$, then the LHS of the inequality is an integer, contradicting the minimality of $p$. Thus, the sum cannot divide $n^2$ if $n$ is composite, as desired. $\blacksquare$

Note that $d_id_{k-i+1} = n$ for all $1 \leq i \leq k$.
Then, $D = d_1d_2 + d_2d_3 + \cdots \ d_{k-1}d_k$ has
\[ D = n^2 (\frac{1}{d_{1}d_{2}} + \frac{1}{d_{2}d_{3}} + \cdots +
\frac{1}{d_{k-1}d_{k}}) \]
Since $1 = d_1 < d_2 < \cdots < d_k$, we have $i < d_i$ for all $1 \leq i \leq k$.
This gives:
\[ \frac{1}{d_{1}d_{2}} + \frac{1}{d_{2}d_{3}} + \cdots +
\frac{1}{d_{k-1}d_{k}}
\leq \sum_{i = 1}^{k-1}\frac{1}{(i)(i+1)}
= \sum_{i = 1}^{k-1}\frac{1}{i} - \frac{1}{i+1}\]
By telescoping, the above sum is equal to $1 - \frac{1}{k} < 1$, so $D < n^2 \cdot 1 = n^2$.

Claim: $D \mid n^2$ if and only if $n$ is prime.\\
It is easy to check that when $n$ is prime, $D = 1 \cdot n = n$, and trivially $n = D \mid n^2$.
Otherwise, assume $n$ is not prime and proceed by contradiction. If $D \mid n^2$, then $Dp = n^2$ for some positive integer $p$. By the above, $p$ cannot be one, so it has to be $\geq$ the smallest non-one divisor of $n^2$. This is the smallest prime in the prime factorization of $n^2$, which is also the smallest prime in the prime factorization of $n$, and therefore $p \geq d_2$.

This means that $d_2d_{k-1}d_k = n^2 \leq Dp$. Equality is only when $d_{k-1}d_k$ is the only term in $D$, i.e. $k = 2$, implying that $n$ is prime, which is a contradiction. Otherwise $n^2 < Dp$, contradicting $Dp \mid n^2$.

By considering the divisors in the form $\frac{n}{d}$ we get it suffices to prove
$$\frac{1}{d_1d_2}+\frac{1}{d_2d_3}+\cdots +\frac{1}{d_{k-1}d_k}\leq 1$$
Now note that
$$\frac{1}{d_1d_2}+\cdots +\frac{1}{d_{k-1}d_k}\leq\frac{d_2-d_1}{d_1d_2}+\cdots \frac{d_k-d_{k-1}}{d_{k-1}d_k}=\frac{1}{d_1}-\frac{1}{d_2}+\cdots +\frac{1}{d_{k-1}}-\frac{1}{d_k}=1-\frac{1}{n}<1$$
For the other part, note that
$$n^2>d_1d_2+\cdots d_{k-1}d_{k}\geq d_{k-1}d_{k}=\frac{n}{d_2}\cdot n=\frac{n^2}{d_2}$$
But $n^2$ and $\frac{n^2}{d_2}$ are the bigger and the second bigger divisors of $n^2$, and the equality is satisfied only when $n$ is prime

[hide]
First we prove that $s < n^2$.

Note that $s = n^2(\sum_{i=1}^{k-1}\frac{1}{d_id_{i+1}}) \leq n^2\sum_{i=1}^{n-1}\frac{1}{i(i+1)} = n^2(1-\frac{1}{n}) = n^2 - n < n^2$

Now we claim only primes $n$ work. For primes $n$, the sum $\sum_{i=1}^{k-1}d_id_{i+1} = n$ which divides $n^2$. If $n$ is not prime, then the sum has more than $1$ term and is larger than $\frac{n^2}{p}$ where $p$ is the smallest prime factor of $n$. Since $\frac{n^2}{p}$ is the largest proper divisor of $n^2$, it follows that $s \nmid n^2$ if $n$ is not prime.
[/hide]


Please slove the following:

1. Find the coordinates of the foot of perpendicular from (1,3) to the line 6x-8y-15=0. Also find the equation of this perpendicular line.

2. From a group of 10students, 6 are to be chosen for a picnic party. There are 3students who decide that either all of them or none of them will join. In how many ways can the picnic party be chosen?

4. Find 4 numbers in geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

I'd be very grateful to your forum if anyone could solve all the above questions...

PS> [size=75]please solve them within 12 hours as I'm having tests. I'd understand if you just gives solutions instead of both explanation and solutions...[/size]

Before you start you should draw a diagram so its easy to see

[img]http://i17.tinypic.com/4lfd3e1.jpg[/img]

1. $ 8y \equal{} 6x \minus{} 15$

$ y \equal{} \frac {3}{4}x \minus{} \frac {15}{8}$

hence, $ m \equal{} \frac {3}{4}$

Since this line is [b]perpendicular[/b] to the y=mx+b with coordinates $ (1,3)$ then you can say that

$ y \equal{} mx \plus{} b$ has gradient $ (m \equal{} \minus{} \frac {4}{3})$. $ [m_1.m_2 \equal{} \minus{} 1]$

Using, $ y \minus{} y_1 \equal{} m(x \minus{} x_1)$ you can find the equation of the perpendicular line.

$ y \minus{} 3 \equal{} \frac {4}{3}(x \minus{} 1)$ $ \equal{} \equal{} >$ expand and simplify and you win get the equation of the perpendicular line

To find the coordinates of the foot of the perpendicualr use simlutaneuos equations

i.e. $ 8y \equal{} 6x \minus{} 15$
$ y \minus{} 3 \equal{} \frac { \minus{} 4}{3}(x \minus{} 1)$

solve this to find $ (x,y)$

--------

heh doing some last minute study?

okay, number 4:

Let $ b$ be the first term and $ r$ be the ratio.
So the progression is:
$ b, br, br^2, br^3, br^4...$
We are given:
$ br^2 \minus{} b \equal{} 9$
$ br \minus{} br^3 \equal{} 18$

$ b(r^2 \minus{} 1) \equal{} 9$
$ br(1 \minus{} r^2) \equal{} 18$

from this we can see that $ \boxed{r \equal{} \minus{} 2}$ (it's negative because the terms in the bracket are flipped, and it's 2 because 9*2 = 18)

pluging it in  $ b(r^2 \minus{} 1) \equal{} 9$ gives $ b(3) \equal{} 9$
Thus $ \boxed{b \equal{} 3}$

So the progression is

$ \boxed{3, \minus{} 6, 12, \minus{} 24}$

Hope this helps  :)

That was really helpfull. Thanks a lot.  :)

But what about my 2nd question? Have anyone figured it out? I can't get it.  :(

[b]2.[/b]

Without the restriction: $ \binom{10}{6} \equal{} \frac{10!}{6!\cdot 4!} \equal{} 210$ possible combinations.

To apply the 3-student restriction we can divide into two possible types:
[list][*]all three “linked” students go, leaving us with a choice of the three remaining students amongst seven: $ \binom{7}{3} \equal{} 35$ combinations;
[*]none of the three “linked” students go, leaving us with a choice of all six students amongst the usual seven: $ \binom{7}{6} \equal{} 7$ combinations.[/list]Adding up, $ 35 \plus{} 7 \equal{} \boxed{42}$ combinations.


search
quadratics
modular arithmetic
number theory unsolved
number theory

a)  Prove that 3 is a primitive root mod $ p$ if $ p$ is a prime of the form $ 2^n \plus{} 1$, $ n > 1$.

b) Prove that 2 is a primitive root mod $ p$ if $ p$ is a prime of the form $ 4q \plus{} 1$, where $ q$ is an odd prime.

a) [url]http://www.mathlinks.ro/viewtopic.php?search_id=846925246&t=237294[/url]
b) The order of $ 2$ divides $ \phi(p) \equal{} 4q$. To show that $ 2$ is a primitive root, it is enough to role out the possibilities that $ 2$'s order is $ 1$, $ 2$, $ 4$, $ q$, or $ 2q$. Actually, it is enough to check $ 4$ and $ 2q$.
The first is easy to eliminate - $ 2^4 \minus{} 1 \equal{} 15 \equal{} 3 \cdot 5$. $ 3$ and $ 5$ are not of the required form.
The second one follows from noticing that $ 2$ is not a quadratic residue mod $ p$ and thus $ 2^{2q} \equiv \minus{} 1 \pmod p$.
(use $ \left( \frac {2}{p} \right) \equal{} ( \minus{} 1)^\frac {p^2 \minus{} 1}{8}$)


topology
advanced fields
advanced fields unsolved

Let X be an infinite set.
Let X be Hausdorff
Let H the topology consisting of all sets A satisfing either A=the empty set
or A complement is at most countable

Show X is at most countable
Show X is Discrete

This is nonsense as stated. What's the real statement?

Let X be an infinite set.
Let X be Hausdorff
Let H the topology consisting of all sets A satisfing either A=the empty set
or A complement is finite or at most countable

Prove the following equivalent

X is at most countable
X is Discrete

I believe the question is asking about what happens if the cocountable topology is Hausdorff.  If so, the sapce is countable and the topology is discrete.

The equivalence between
X is at most countable
X is Discrete
is true in every space with cocountable topology :

1) If X is at most countable, every subset of X is open, so {x} is open, so it's discrete
2) If X is discrete, $X\setminus \{x\}$ is closed (that is, at most countable) for every $x\in X$, so $X$ is at most countable.

For the original problem, if you suppose X has the cocountable topology and is Hausdorff, then for every $x,y \in X$ there exist $U,V\subset X$ open such that $x\in U$, $y\in V$ and $U\cap V=\emptyset$, that is $U\subseteq V^C$; but $V^C$ is at most countable, so $U$ is at most countable and also $U^C$ is. Therefore $X=U\cup U^C$ is at most countable.
This means that $X$ is discrete and if you suppose it has infinite elements, it is countable.

This result shows that do not exist spaces with the requested properties, because in every Hausdorff space the points are closed, so that it cannot be discrete.


[hide] Say that $\frac{1}{x-2}=a$ and $\frac{1}{y-3}=b$.  Thus,
\[2a+3b=-7 \]
and
\[5a+2b=-1 \]
Solving the pair of linear equations, we find that $a=1, b=-3$  So, $\frac{1}{x-2}=1\Rightarrow x-2=1\Rightarrow \boxed{x=3}$ and $\frac{1}{y-3}=-3\Rightarrow 9-3y=1\Rightarrow \boxed{y=\frac{8}{3}}$. [/hide]

[quote="Ignite168"]Find the ordered pair (x,y) that satisfies:
$\frac{2}{x-2}+\frac{3}{y+3}=-7$
$\frac{5}{x-2}+\frac{2}{y+3}=-1$[/quote]

[hide="solution"]Multiply the first equation by 2 and the second by 3 and subtract the two.

$\frac{4}{x-2}-\frac{15}{x-2}=-11$

$\frac{1}{x-2}=1$

$x=3$

Plug this back into the first equation to get $2+\frac{3}{y+3}=-7$ so $y=-\frac{10}{3}$

$\boxed{\left(3,-\frac{10}{3}\right)}$[/hide]


i voted for a bunch of bananas

MORE IS ALWAYS BETTER

Wow post #1367, I honestly admit I am spamming. I deserve points for honesty.

Ok, banana fun is over. Everyone back to his forum :)


probability
real analysis
real analysis unsolved

Is the set of fractional parts of the harmonic sums dense in $ [0,1]$ (joke)? Is it equidistributed?

The definitions are a little fuzzy, but I'd have to say that it's not equidistributed; if we choose $ n$ uniformly from $ \{1,2,\cdots,N\}$ and take the $ n$th partial sum of the harmonic series, its fractional part will be biased toward the fractional parts for $ n$ close to $ N$. As $ N\to\infty$, the density at $ x$ looks like $ \frac{e^{-[c-x]}}{1-e^{-1}}$, where $ [$ $ ]$ is the fractional part and $ c$ is the fractional part of the sum from 1 to $ N$.

The answer would be completely different if we looked at other probability distributions. There is no canonical choice, after all.

I would say no:
$ S_{n}=1+\frac{1}{2}+\ldots+\frac{1}{n}\approx \ln(n)+\gamma+o(1)$, and define $ A(n)$ and $ B(n)$ the smallest integers such that $ S(A(n))>n$ and $ S(B(n))>n+\frac{1}{2}$. Then $ A(n) \approx C \exp(n)$ and $ B(n)-A(n) \approx C(e^{\frac{1}{2}}-1) \exp(n)$. Suppose that the sequence $ S(n)$ is equidistributed modulo $ 1$. Then $ D(n) = \frac{Card(k \leq n : \{S_{k}\}\in [0;\frac{1}{2}])}{n}\to \frac{1}{2}$. But:
$ D(A(n)-1) = \frac{1}{A(n)-1}[(B(1)-A(1))+(B(2)-A(2))+\ldots+(B(n-1)-A(n-1))]$ hence $ D(A(n)) \to (e^{\frac{1}{2}}-1) (e^{-1}+e^{-2}+e^{-3}+\ldots) \neq \frac{1}{2}$


Ok, I have a question...how do you box answers? Also, how to I change my hidden post to say like.."Solution" rather than "Click here to reveal content"? Thanks.

The code is \boxed{}, so you get $ \boxed{5}$.

When it says hide, add an equal sign and then the text, so its says hide=Solution.

Yes. After the first hide inside the brackets, put "=blah blah blah" without the quotes. Note: the blah blah blah should also be inside the brackets.

Example.

[code][hide=put your text here!]
Ooooh, hidden![/hide][/code]

[hide="put your text here!"]
Ooooh, hidden![/hide]


pigeonhole principle
combinatorics proposed
combinatorics

Let $A=\{a_{1}a_{2}\cdots a_{k}\,|\,k\in \mathbb{N},a_{i}=0,1\}\subset \mathbb{N}.$ By pigeonhole principle prove that for each $n\in \mathbb{N}$ there exists $a\in A$ such that $a=mn$ (for some $m\in \mathbb{N}$).

1. Problem 23 of what?
2. This question and equivalent variants have appeared many times on the forum.  I recommend a search.

[quote="JBL"]
2. This question and equivalent variants have appeared many times on the forum.  I recommend a search.[/quote]

I searched but not found. Please help.

These two threads have closely related questions:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12239
http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=4248

take the numbers of form $\frac{10^{n}-1}{9}$

two of this are equals (mod. n )

so, your difference are one multiple of n!  :)


geometry
circumcircle
incenter
IMO Shortlist
reflection

Given an acute triangle $ ABC$ with $ \angle B > \angle C$. Point $ I$ is the incenter, and $ R$ the circumradius. Point $ D$ is the foot of the altitude from vertex $ A$. Point $ K$ lies on line $ AD$ such that $ AK \equal{} 2R$, and $ D$ separates $ A$ and $ K$. Lines $ DI$ and $ KI$ meet sides $ AC$ and $ BC$ at $ E,F$ respectively. Let $ IE \equal{} IF$.

Prove that $ \angle B\leq 3\angle C$.

[i]Author: Davoud Vakili, Iran[/i]

We begin with a preliminary result. We will work with unoriented angles, i.e. the solution is configuration-dependent.

[b]Lemma 1[/b]. The magnitude of angle $ \angle{KID}$ is $ \frac {B - C}{2}$.

[i]Proof of Lemma 1[/i]. We will give a metric proof different from the official ones.
Denote by $ A^{\prime}$ the antipode of $ A$ with respect to the circumcircle $ (O)$ of triangle $ ABC$. In this case, triangle $ AKA^{\prime}$ is isosceles and since the cevians $ AH$ and $ AO$ are isogonal with respect to the angle $ \angle{BAC}$ we deduce that $ I$ lies on the line bisector $ AM$ of the segment $ A^{\prime}K$ and moreover, the midpoint $ M$ of $ A^{\prime}K$ is the midpoint of the small arc $ BC$.
Since $ \angle{DAI} = \angle{DAO}/2 = (B - C)/2$, it is suffice to prove that $ \angle{KID} = \angle{DAI}$, which is equivalent with the fact that the line $ KI$ is tangenct to the circumcircle of triangle $ DAI$ at $ I$. In this case, we will resume to proving that $ KA \cdot KD = KI^{2}$, which will yield our conclusion.
As I was saying,
$ KI^{2} = AK^{2} + AI^{2} - 2AI \cdot AK \cdot \cos{\frac {B - C}{2}}$
$ = 4R^{2} + \frac {r^{2}}{\sin^{2}{\frac {A}{2}}} - \frac {4Rr\cos{\frac {B - C}{2}}}{\sin{\frac {A}{2}}}$
$ = 4R^{2} + \frac {r^{2}bc}{(p - b)(p - c)} - \frac {4Rr(b + c)}{a}$
$ = 4R^{2} + \frac {bc(p - a)}{p} - \frac {4Rr(b + c)}{a}$
$ = 4R^{2} - \frac {4pRr(b + c) - abc(p - a)}{ap}$
$ = 4R^{2} - \frac {4RS}{a}$
$ = 2R(2R - h_{a}) = KA \cdot KD$. This proves Lemma 1.

We will end this proof apparently in the same manner as the official proof. Denote by $ A_{1}$, $ B_{1}$ the tangency points of the incircle with the sides $ BC$ and $ CA$, respectively. Since the two triangles $ IEB_{1}$ and $ IFA_{1}$ are congruent, $ \angle{IEB_{1}} = \angle{IFA_{1}}$. Now since $ \angle{B} \geq \angle{C}$, $ A_{1} \in (CD)$ and $ F \in A_{1}D$. Thus, $ \angle{IFC}$ is acute. We distinguish two cases, depending on the position of $ B_{1}$ and $ E$ on the side $ AC$.

(1) $ A - B_{1} - E - C$: We have that $ \angle{IFC} = \angle{IEA}$, and thus the quadrilateral $ CEIF$ is cyclic which implies that $ \angle{FCE}( = \angle{C}) = \angle{KID} = (B - C)/2$ and so $ B = 3C$.

(2) $ A - E - B_{1} - C$: In this case $ CEIF$ is a deltoid such that $ \angle{IEC} = \angle{IFC} < \pi/2$. Then we have $ \angle{FCE}( = \angle{C}) > 180^{\circ} - \angle{EIF} = \angle{KID} = (B - C)/2$. Hence, $ \angle{B} < 3\angle{C}$. This proves our problem.

Two more proofs of [b]pohoatza[/b]'s lemma:
[hide="Proof 1"]
Let $ O$ be the circumcenter of $ \triangle ABC$, $ M$ be the midpoint of arc $ BC$ of $ (O)$, $ B'$ be the reflection of $ B$ over the angle bisector of $ A$, and define $ C'$ to be that of $ C$.  Let $ I_a$ be the $ A$-excenter of $ \triangle ABC$.  We have that since $ AD$ is the reflection of $ AO$ over the angle bisector of $ \angle BAC$, $ AK$ is the diameter of $ \triangle AB'C'$ (note that $ AK \equal{} 2R)$.  Now, since $ M$ lies on the $ (O)$ and on the angle bisector of $ \angle BAC$, it must lie on the circumcircle of $ \triangle AB'C'$, establishing that $ \angle KMA \equal{} 90$.  Since $ M$ is the midpoint of $ II_a$, we have that $ KI_a \equal{} KI$.  Furthermore, we have that $ \angle KDB \equal{} 90 \equal{} \angle KC'B$, so $ BC'KD$ is cyclic.  We may also say that since $ MI \equal{} MI_a \equal{} MB \equal{} MC \equal{} MC'$, $ BC'I_aI$ is cyclic.  This means that $ AD\cdot AK \equal{} AB\cdot AC' \equal{} AI\cdot AI_a$, meaning that $ DII_aK$ is cyclic.  Thus, $ \angle KI_aI \equal{} \angle KII_a \equal{} \angle KDI_a$, so $ \triangle KDI_a\sim \triangle KI_aA$, meaning that $ KD\cdot KA \equal{} KI_a^2 \equal{} KI^2$, so $ \triangle KIA\sim \triangle KDI$.  Thus, $ \angle KID \equal{} \angle KAI \equal{} 90 \minus{} \frac {(\angle ABC \plus{} \angle ACB)}{2} \minus{} (90 \minus{} \angle ABC) \equal{} \frac {\angle ABC \minus{} \angle ACB}{2}$. [/hide]

[hide="Proof 2"]
Define everything to be the same as before (but we won't use some of the points created).  Let $ AD \equal{} h_a$, $ BC \equal{} a$, $ AC \equal{} b$, and $ AB \equal{} c$.  We have that $ 4R[ABC] \equal{} abc$, so $ 2Rh_aa \equal{} abc$, so $ 2Rh_a \equal{} bc$, meaning that $ AK\cdot AD \equal{} AB\cdot AC$.  Yet, we may also say that since $ IBI_aC$ is cyclic, we have that $ \angle AI_aC \equal{} \angle IBC \equal{} \angle ABI$ and $ \angle I_aAC \equal{} \angle BAI$.  This means that $ \triangle AIB\sim \triangle ACI_a$, so $ \frac {AB}{AI} \equal{} \frac {AI_a}{AC}$, so $ AI\cdot AI_a \equal{} AB\cdot AC \equal{} AD\cdot AK$, meaning that $ DII_aK$ is cyclic and we proceed as we did above. [/hide]

let$ I_{a}$ is $A-$excenter.
let $AI$ meet $BC$ at $S$.
$AI.AI_{a}=AB.AC=2R.AD=AD.AK$
$\widehat{KII_{a}}=\widehat{I_{a}DK}.(A,S,I,I_{a})=-1\Rightarrow \widehat{KDS}=\widehat{IDS}=90-\widehat{FII_{a}}$
case 1: $\widehat{IFC}=\widehat{IEC}$ easy by angle chasing to find $B=3C$
case 2: $\widehat{IFC}+\widehat{IEC}=180$ easy by Calculation  angle $\widehat{DKI}$

I give a proof of the fact that $\angle KID=\frac{\angle B-\angle C}{2}$ and the rest is the same as the proof given by Pohoatza.

Let $A'$ be the antipode of $A,$ $J$ be the $A$-excenter for triangle $ABC$ and $M$ the midpoint of $IJ$. Note that $\angle AMA'=90^{\circ}$ and from the fact that $\triangle AIC \sim \triangle ABJ$ we have $AI\cdot AJ=AB\cdot AC$.

Also, $AD\cdot AK=AD\cdot AA'=AB\cdot AC$ where the last equality follows since $\triangle ABD \sim \triangle AA'C$.

Finally, as \begin{align*} KD\cdot KA=AK^2-AK\cdot AD=AA'^2-AI\cdot AJ &= AA'^2-(AM^2-MI^2) \\
&=(AA'^2-AM^2)+MI^2\\
&=A'M^2+MI^2=A'I^2=KI^2 \end{align*} we get $\triangle KID \sim \triangle KAI \Longrightarrow \angle KID=\angle DAI$ and the result follows.

[b]Note[/b] The most difficult part of this problem was to actually "guess" the result we showed here since it is not a very "well-known" statement.

Notice that $AD$ lies on the same side of the angle bisector as $B$.So upon assuming that $\angle IFC$ is obtuse we obtain that $\measuredangle(AI,BC)$ is acute a contradiction and hence $\angle AFC$ is acute.

Let $I_a$ be the excenter ,$A'$ the antipode.Notice tha $\sqrt{bc}$ inversion $AA'\cdot AD=bc=AD\cdot AK=AI\cdot AI_1$ $\implies$ $IDKI_A$ is cyclic and as $KI=KI_A$ upon inverting thru $K$ with radius $KI$ we obtain $KI^2=KD\cdot KA$ $\implies$ $KI$ is tangent to $\odot AID$ $\implies$ $\angle DIK=\angle DAI=\frac{\beta-\gamma}{2}$.

If one of $\angle CEI,\angle CFI$ is obtuse ,by first paragraph, other one is acute $\implies$ in this case $AIFE $is cyclic and so $\gamma=\frac{\beta-\gamma}{2}$.So assume they're both acute (and hence equal):
$$\angle CDI=\angle CFI-\angle FID=\pi-\gamma-\angle CFI  \implies \gamma=\pi-2\angle CFI+\angle FID$$
 $\gamma\geq \frac{\beta-\gamma}{2}$ and hence the desired.$\blacksquare$

[hide=Diagram][img width=50]https://media.discordapp.net/attachments/986049049834713139/1161833666931073105/Screenshot_2023-10-11_at_18.12.57.png?ex=6539bcec&is=652747ec&hm=61c8becb91dcf5a1387da40b89048f0f45b3450d0d6ec96e7f2a1a771479b748&=&width=1328&height=1144[/img][/hide]

Let $A'$ be the $A$-antipode. Clearly, since $AK=AA'$, and $AI$ is the angle bisector of $\angle DAO$, when $AI$ intersects $KA'$ at $L$, $AL$ is also the altitude. Thus, $L$ lies on $(ABC)$. In fact, it is the arc midpoint of $BC$. Thus, by incenter-excenter lemma, $L$ is the center of a circle $(BIC)$. Since $\triangle ABD\sim \triangle AA'C$, we have $AD\cdot AA'=AB\cdot AC$. Let $B'\neq C$ be the point of intersect of circle $L$ with $AC$. Clearly, $AB=AB'$. We have
\[KI^2 = A'I^2 = A'A^2 - AL^2 + IL^2 = A'A^2 - \text{Pow}_{(BIC)}(A) = A'A^2 - AB\cdot AC\]
On the other hand,
\[KA\cdot KD=KA^2 - AK\cdot AD=A'A^2-AA'\cdot AD=A'A^2-AB\cdot AC\]
so $KI^2=KA\cdot KD$. We have $\angle KID=\angle KAI=\tfrac{\angle B - \angle C}{2}$. Let $H_A$ and $H_B$ be the feet of the altitudes from $I$ to $BC$ and $AC$, respectively. Note that $F$ must be on the segment $BH_A$. Let $E'$, regardless of the position of $E$, be the point on segment $CH_B$ for which $IE'=IF$. Evidently, $IF=IE'$ implies $\angle H_BIE'=\angle H_AIF$ so $\angle H_BIH_A=\angle FIE'=180^\circ-\angle C$. Thus,
\[\angle DIF=180^\circ-\angle FIE\ge 180^\circ-\angle FIE'=\angle C\]
and we are done.


function
modular arithmetic
induction
algebra unsolved
algebra

Define the function $f: \mathbb N \cup \{0\} \to \mathbb{Q}$ as follows: $f(0) = 0$ and \[ f(3n+k) = -\frac{3f(n)}{2} + k ,  \] for $k = 0, 1, 2$. Show that $f$ is one-to-one and determine the range of $f$.

Wouldn't this imply the existence of a bijection from the reals to the rationals?   :|

I've modified the statement of the problem. It's probably the intended statement (of course there is no one-to-one map from the nonnegative reals to the rational numbers :) ).

As pointed out earlier,

\[f(\sum_{n=0}^{m} a_{n}3^{n}) = \sum_{n=0}^{m}a_{n}(\frac{-3}{2})^{n})\]

where $a_{n}\in \{0,1,2\}$. Now suppose that this mapping is not one-to-one, then

\[\sum_{n=0}^{m}a_{n}(-1.5)^{n}= \sum_{n=0}^{m}b_{n}(-1.5)^{n}\]

or

\[\sum_{n=0}^{m}(a_{n}-b_{n}) (-1.5)^{n}= 0\]

so there is some non-zero weighting that causes the sum to come out to zero. Now, multiply by $2^{m}$, and let $d_{n}= a_{n}-b_{n}$, then we get

\[\sum_{n=0}^{m}d_{n}(-3)^{n}2^{m-n}= 0\]

and taking this in mod $3$ we get $d_{0}\equiv 0$, so $d_{0}= 0$. Now factor out a $(\frac{-3}{2})$ from the sum, and by the same logic we get $d_{1}= 0$, etc., so that $d_{i}= 0$ for all $i$ and the function is one-to-one.

We claim the range is all fractions of the form $\frac{n}{2^{m}}$. Note that all fractions not of this form are unattainable, since only two can divide the denominator of any value of this function. We start by showing that all integers are attainable, for, given $n$, we can get both $n+1$ and $n-1$, using the identities $2*(\frac{-3}{2})^{2}+1*(\frac{-3}{2})+0 = 3$ and $2*(\frac{-3}{2})+3 = 0$ to carry, treating the addition as addition in base $\frac{-3}{2}$. Since we have $0$, by this observation we have all the integers.

Now, suppose we have all multiples of $\frac{1}{2^{n}}$. Then we have $\frac{k}{2^{n}}$ for all $k$, and so also $\frac{-3k}{2^{n+1}}+\{0,1,2\}$ for all $k$. But we can always satisfy the equation

\[\frac{m}{2^{n+1}}= \frac{-3k}{2^{n+1}}+\{0,1,2\}\]

as it is equivalent to

\[m =-3k+\{0,2^{n+1},2^{n+2}\}\]

or

\[m \equiv \{0,1,2\}\pmod{3}\]

which is of course always true. We therefore have all fractions of the form $\frac{m}{2^{n}}$, as stated.

This problem seems too simple to be Q6 in USA TST  :maybe:
Here's a slightly different approach $ f(n)$ concisely:
Suppose $ f$ is not injective. Then for some $ a \neq b$, we have $ f(a) \equal{} f(b)$. Pick the pair $ (a,b)$ with smallest value of $ min (|a|, |b|)$, such that $ a \neq b$ and $ f(a) \equal{} f(b)$. Let $ a \equal{} 3s\plus{}t, b\equal{}3u\plus{}v$ where $ t,v$ are $ 0,1$ or $ 2$.
We have $ \frac{\minus{}3}{2} f(s) \plus{} t \equal{} \frac{\minus{}3}{2} f(u) \plus{} v$. If $ u\equal{}v$, then $ f(s)\equal{}f(u)$, and $ min (|s|, |u|) < min (|a|, |b|)$, which contradicts the minimality of the pair $ (a,b)$.
If $ u \neq v$, we have $ f(s) \minus{} f(u) \equal{} \frac {\minus{}2}{3} (v\minus{}t)$. But we can easily see, from the condition that the range of $ f(n)$ is a subset of all rationals, with denominator a power of $ 2$, and so $ f(s) \minus{} f(u)$ is also is a rational, with denominator a power of $ 2$. Since $ 3$ does not divide $ v\minus{}t$ (since $ v\minus{}t \neq 0$), we know that $ \frac {\minus{}2}{3} (v\minus{}t)$ is a fraction, that when reduced, has a denominator divisible by $ 3$, which is a contradiction.
Next we prove that all integers $ n$ are part of the range of $ f$: By some simple calculations we can see that small integers, $ (\minus{}4, \minus{}3 ... 3)$ are part of the range. Now suppose there exists an integer $ n$ that is not part of the range of $ f$. Pick the integer with smallest value of $ |n|$ which is not part of the range of $ f$. Let $ n_{1}$ be the residue of $ n$ mod $ 3$ (so $ n_{1}$ is $ 0, 1$ or $ 2$). If $ m \equal{} \frac {\minus{}2}{3} (n\minus{}n_{1})$ is part of the range of $ f$ suppose $ f(t) \equal{} \frac {\minus{}2}{3} (n\minus{}n_{1})$, then we have $ f(3t\plus{}n_{1}) \equal{} n$, which is a contradiction since $ n$ cannot be expressed such. Thus $ m \equal{} \frac {\minus{}2}{3} (n\minus{}n_{1})$ is also not part of the range of $ f$. But $ |m| < |n|$ (since $ n$ is not small: we have shown small integers, $ \minus{}4... 3$ are part of the range of f), which contradicts the minimality of $ |n|$.
Now we can prove by induction, that if all rationals of form $ \frac {k}{2^{l}}$ are part of the range of $ f$, then all rationals $ \frac {k}{2^{l\plus{}1}}$ can be expressed such. We have proven this for $ l\equal{}0$ above. Now for some rational $ \frac {k}{2^{l\plus{}1}}$: first choose $ t$ so that $ 3 | k \minus{} t*2^{l\plus{}1}$, and $ t \equal{} 0, 1$ or $ 2$. By induction, there exists $ s$ so that $ f(s) \equal{} \minus{}  \frac { k \minus{} t*2^{l\plus{}1}}{3*2^l}$, so choose $ u \equal{} 3s\plus{}t$ and we can compute that $ f(u) \equal{} \frac {k}{2^{l\plus{}1}}$, as required.
Since it's clear that all numbers in range of $ f$ are of the form $ \frac {k}{2^l}$, and we have proved all such numbers are part of the range, the range consists of those numbers.


[quote="rcv"]Two cubes have edge lengths of 4 and 6 inches.  What is the ratio of the surface area of the smaller cube to the surface area of the larger cube.  Express your answer as a common fraction.[/quote]

[hide]
Formula for the surface area of a cube:
6s 2

6*4 2 = 6*16 = 96

6*6 2 = 6*36 = 216

96/216 = 4/9
[/hide]
 :)

Good answer.

Just quick note:

[hide="IF"]If the problem said the edge was 999 and 9999, squaring and then multiplying by 6 is not that nice thing to do.

In this problem, it didn't matter but here is easier way, basically same but cancels out so you have less multplication.

Call first cube S and second one s.

First cube's volume is S^2*6

Second cube's volume is s^2*6

First volume: Second volume is equal to:

\[\frac {S^2 \cdot 6}{s^2 \cdot 6}\]

\[\frac {S^2}{s^2}\]

So instead of multiplying 16*6 and 36*6 and set 96/216, you could just had 16/36 and get the same answer.

[/hide]


Let $ a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n$ be real numbers with
$ a^2_1 \plus{} a^2_2 \plus{} \cdots \plus{} a^2_n \equal{} b^2_1 \plus{} b^2_2 \plus{} \cdots \plus{} b^2_n \equal{} 1,$
and
$ a_1b_1 \plus{} a_2b_2 \plus{} \cdots \plus{} a_nb_n \equal{} 0.$
Prove that
$ (a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)^2 \plus{} (b_1 \plus{} b_2 \plus{} \cdots \plus{} b_n)^2 \le n.$

hello, it must be
$ \left(\sum_{i\equal{}1}^na_i\right)^2\plus{}\left(\sum_{i\equal{}1}^nb_i\right)^2\le n$
Sonnhard.

if i didn't make any mistake expand that
$ \sum^{n}_{j\equal{}1}(1\minus{}(\sum^{n}_{i\equal{}1}a_{i})a_{j}\minus{}(\sum^{n}_{i\equal{}1}b_{i})b_{j})^{2}\geq0$
SAIF :)

[quote="saif"]if i didn't make any mistake expand that
$ \sum^{n}_{j \equal{} 1}(1 \minus{} (\sum^{n}_{i \equal{} 1}a_{i})a_{j} \minus{} (\sum^{n}_{i \equal{} 1}b_{i})b_{j})^{2}\geq0$
SAIF :)[/quote]
This is a nice proof, but I wonder are there any other proofs?


Does anyone here have an imvu account? If you want to imvu chat with me my account is asmallgoat. Plus I have an awesome pool. :lol:

http://www.imvu.com iming with 3d characters so while you chat you can beat people up and stuff..

Oh that sounds wonderful I'll go get one.

A bit like second life.
Anyone I registered as zedboy.
Downloading now ...


Does anyone know if a 5th grader can take the middle school level Olympiads if they are ready?

I don't see why not.  We had a 5th grader taking the AIME, as I recall...

[quote="kstan013"]Well...yes...but the key word is [b]middle school olympiads[/b] (ok, word[b]S[/b] then :P ). The rules for MOEMS may be different from those of the AMC contest. I'm not sure if they are, but just a possibility to consider. For example, MATHCOUNTS does not allow anyone under 6th grade... :maybe:[/quote]
Umm...your wrong there kstan. Mathcounts does. You should see my team the past two years. If I remember correctly(which I probably don't) you are able take it when you are younger...

[quote="mathmom"]I am the PICO for the MOEMS olympiads at our school, and also the MATHCOUNTS coach.

MOEMS does allow students in lower grades to take the contests if they want.  At [url]http://www.moems.org/contests.htm[/url] they state:

[quote]While younger students will be accepted into either division, we do not recommend it. Without proper support, younger students can easily become afraid of math, rather than excited by it.[/quote]

MATHCOUNTS does not officially allow this, though some districts might allow it unofficially at chapter.  At [url]http://www.mathcounts.org/webarticles/anmviewer.asp?a=97[/url] they state:

[quote][b]Students enrolled in the 6th, 7th or 8th grade are eligible to participate in MATHCOUNTS competitions. [/b]Students taking middle school mathematics classes who are not full‑time 6th, 7th or 8th graders are not eligible.[/quote][/quote]
well, i think that mathcounts rules don't quite matter unless say you make it to nats

[quote="funcia"]
well, i think that mathcounts rules don't quite matter unless say you make it to nats[/quote]

I know our chapter has allowed lower grade kids to compete unofficially as individuals in the past, but not on a school's official team.  I don't think they'd allow such a student to win a trophy, or a place at State.  This year, they did not allow them at all.  Our state has been known to let an underage kid write the contest "just for fun" if there are extra seats due to someone not showing up, but their scores do not count for trophies or qualifying to nats, and likewise the team they did team round with could not win a trophy either.  This is up to the discretion of the State or Region coordinator, I believe, but I do think MATHCOUNTS frowns upon it, and would not approve of a state or region allowing an underage competitor to advance.


linear algebra
matrix
function
linear algebra unsolved

Let be $ A \in M_n(R)$ $ A_* \equal{} [A_{ij} ]_{i,j \equal{} \overline {1,n} }^{t}$,profe that if $ detA \geq 0$ then
$ det(A \plus{} xA_*) \geq 0 ,\forall x\geq 0$.

Assume $ A*$ is the transpose of the matrix of $ A$-cofactors.
Case 1: $ A$ is invertible. Then $ A*\equal{}det(A)A^{\minus{}1}$. $ det(A\plus{}xA*)\geq{0}$ iff $ det(A^2\plus{}xdet(A)I)\geq{0}$ .This expression is in the form $ det(A^2\plus{}B^2)$ where $ A,B$ commute and are real matrices. We conclude using $ A^2\plus{}B^2\equal{}(A\plus{}iB)(A\minus{}iB)$.
Case 2:$ A$ is singular. We conclude using the following points:
1) There xists a sequence of real invertible matrices $ (A_k)_k$ that converges to $ A$.
2) The functions $ A\rightarrow{det(A)}$ and $ A\rightarrow{A*}$ are continuous.


Can we take both the AIME I and AIME II? Are these AIME's like the A and B AMC's?

Okay, so I may be a little bit out of date, but 2 years ago, this is what was true:

The AIME 2 was very much [b]not[/b] equivalent to the AMC B.  Rather, it was an exam designed to be taken only by those individuals who absolutely could not take the AIME at its regular time.  (For example, my junior year I was in Cambridge for a week-long program and so I took the AIME 2 when I returned.)  Under no circumstances could anyone take both the AIME and the make-up exam.  So, based on that, you should still treat the AIME as a 1-time exam, which happens to have a make-up if people really need it.  You should not treat it like the B exam.

It's my understanding that you can only take one of the exams.  In fact it seems that only those that couldn't make the first take the second.  You might want to check out [url=http://www.unl.edu/amc]www.unl.edu/amc[/url]

There was a thread about this last year.  Although AMC director never weighed in himself, you might look it over:

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=AIME+II&t=11392

From the AMC Website:
http://www.unl.edu/amc/d-publication/d1-pubarchive/2004-5pub/05TM1012/html/05tm-05-text.html

"Under no circumstances can a student take both AIME's."

This same statement is on page 9, Section X. of the AMC 10A /AMC 12 A Teachers' Manual, which should be in your teacher's hands already if your teacher ordered materials for the A-date.

Steve Dunbar
AMC Director


Let $ a,d$ be relatively prime nos. [positive integrs}
Show that $ ax\plus{}by\plus{}n$  has non negative integer solutions $ (x,y)$ for all integers $ n>ab\minus{}a\minus{}b$. WHat if $ n\equal{} ab\minus{}a\minus{}b$?

first of all, i believe you meat $ a,b$ in the first line and
$ ax\plus{}by\equal{}n$ in the second line
this may be proven by the frobenius coin theorem or the [url=http://www.artofproblemsolving.com/Wiki/index.php/Chicken_McNugget_Theorem]Chicke Mcugget Theorem[/url]


How many non-overlapping circles of radius 1 can fit inside a circle of radius 3?
I got 7 but apparently I am wrong? Can somebody tell me what the answer is?

it's really not 7? i find that hard to believe...

Eight circles need a big circle with radius 3.3+. So the answer is definitely seven.

Unless "overlap" includes touching edges, in which case the answer is... five?

[quote="math_explorer"]Eight circles need a big circle with radius 3.3+. >>> So the answer is definitely seven. <<<

Unless "overlap" includes touching edges, in which case the answer is... five?[/quote]

No, this does not follow.  The answer may indeed be seven, but from your first sentence that only
shows that seven is an upper bound.  Just because it does not work for eight circles does not
show it works for one fewer circles.

[quote="pluristiq"]the term "non-overlapping" usually allows for tangency...[/quote]

Not only tangency, but also possible is that a side of a figure
(or part of a side of a figure) coincides with the side
(or part of a side) of another figure.  Or more than one side
(or part of one side) of the first figure may be coincident
with one, or more than one, side(s) (or parts of a side) with
another figure.

An example of some of this, but not all aspects/cases of it,
is that two or more overlapping polygons share positive
area as a consequence of their perimeters crossing each
other in at least two places.

However, there could be a way of two or more polygons,
whose perimeters do not cross, but their areas overlap,
as in a triangle inside of a square.

Also, the figures need not be polygons.

It obviously doesn't follow from what I posted, but IMHO fitting 7 unit circles in one with radius 3 is trivial, and from the previous posts I suspect I'm not the only one who thinks so.

All this discussion about overlapping is nice, but is it helping us? Circles do not have sides.

[quote="math_explorer"]

 Circles do not have sides.[/quote]

This is nothing different than part of what I indicated in my post.
Circles not having sides doesn't keep them from being able to
overlap each other; they can share positive area.
Except in the problem, they are stated as non-overlapping.

And, seven circles sounds like the correct answer to me, also,
from the proposer's question.


function
number theory
greatest common divisor
totient function
number theory unsolved

I am reading through a book and it says $ n \equal{} \sum_{d|n}(\frac {n}{d}) \equal{} \sum_{d|n} \phi(n)$
But..this is wrong. A simple case, take $ n \equal{} 8$. $ \sum_{d|n}(\frac {n}{d}) \equal{} 4 \plus{} 2 \plus{} 1 \neq 8$ where I used $ d \equal{} \{ 2,4,8\}$

Am I doing something basic very wrong? =)

The set of divisors includes $ 1$.  That identity's still quite wrong, though.  The correct statement is that $ \sigma_1(n) \equal{} \sum_{d | n} d \equal{} \sum_{d | n} \frac {n}{d}$, although the first and third expressions are in fact identical.

What does the subscript on $ \sigma$ mean? The $ \sigma_1(n)$.

Edit: And are you using $ \sigma$ as the totient function as well?

t0rajir0u used $ \sigma$ as the sum of divisors function and $ \sigma_x(n)$ is the sum of the xth powers of the divisors of n

We will show that $ \sum_{d \mid n} \phi(d) \equal{} n$

[u][i]Proof[/i][/u]:

Let $ S$ denote the set $ \{1,2,\dots , n \}$.  We distribute the integers of $ S$ into disjoint sets as follows.  For each divisor $ d$ of $ n$, let

$ A(d) \equal{} \{ k : (k,n) \equal{} d, 1\le k \le n \}$.

That is, $ A(d)$ contains those elements of $ S$ which have the gcd $ d$ with $ n$.  The sets $ A(d)$ form a disjoint collection whose union is $ S$.  THerefore if $ f(d)$ denotes the number of integers in $ A(d)$, we have

$ \sum_{d \mid n} f(d) \equal{} n$

But $ (k,n) \equal{} d$ if and only if $ (k/d,n/d) \equal{} 1$ and $ 0 < k \le n$ if and only if $ 0 < k/d \le n/d$.  Therefore, if we let $ q \equal{} k/d$ there is a 1-1 correspondence between the elements in $ A(d)$ and tose integers $ q$ satisfying $ 0 < q \le n/d,(q,n/d) \equal{} 1$.  THe number of such $ q$ is $ \phi (n/d)$.  Hence $ f(d) \equal{} \phi (n/d)$ and thus we have

$ \sum_{d \mid n} f(d) \equal{} \sum_{d \mid n} \phi (n/d) \equal{} n$

But, $ \sum_{d \mid n} \phi (n/d) \equal{} \sum_{d \mid n} \phi (d)$, so we are done.

also

for $ n\equal{}8$, $ d \mid n \equal{} \{1,2,4,8 \}$

therefore $ \sum_{d \mid n} \phi(d) \equal{} \phi (1) \plus{} \phi(2) \plus{} \phi(4) \plus{} \phi(8) \equal{} 1 \plus{} 1 \plus{} (2^2\minus{}2) \plus{} (2^3 \minus{} 2^2) \equal{} 1 \plus{} 1 \plus{} 2 \plus{} 4 \equal{} 8$


Πώς μπορούμε να δείξουμε ότι το σύνολο $ \{(x,y)\in \mathbb R^2 |1 \leq x^2\plus{}y^2 \leq 4\}$ είναι συνεκτικό;

[quote="giannis18"]Πώς μπορούμε να δείξουμε ότι το σύνολο $ \{(x,y)\in \mathbb R^2 |1 \leq x^2 \plus{} y^2 \leq 4\}$ είναι συνεκτικό;[/quote]

:huh:

Ok, καταλαβαίνω ότι δε σου αρκεί ότι είναι εποπτικά προφανές, οπότε δοκίμασε κάτι από τα παρακάτω:
1. Πάρε δύο αυθαίρετα σημεία του συνόλου σου και φτιάξε μία συνεχή καμπύλη που να τα ενώνει (ένα τόξο κύκλου και μία ευθεία αρκούν). Path-connected $ \implies$ connected, hence...
2. Το ορθογώνιο $ R\equal{}[0,2\pi]\times[1,4]$ είναι συνεκτικό και η συνάρτηση $ f: R\to\mathbb{R}^2$ με $ f(r,\theta) \equal{} (r \sin\theta, r \cos \theta)$ είναι συνεχής. Ο δακτύλιος $ f(R)$ είναι το σύνολο σου, οπότε τελείωσες (continuous image of a connected set and all that).

Πάντως, no offense και [i]εντελώς[/i] φιλικά, ας προσπαθήσουμε να μη βάζουμε προβλήματα που είναι απλές εφαρμογές των ορισμών ή που δεν επιδεικνύουν κάποιο ενδιαφέρον φαινόμενο... αλλιώς θα καταλήξουμε να βάζουμε εδώ ολοκληρώματα σαν το $ \int \ln (1\plus{}x^2) dx$ μόνο και μόνο επειδή τα συναντάει κανείς σε κάποια Ανάλυση Ι... :|

Cheerio,

Durandal 1707

No offence taken :lol: και ευχαριστώ για την απάντηση :)

Να προσθέσω μόνο κάτι που συνήθως δεν το λέμε αλλά πιστεύω είναι αρκετά σημαντικό. Πολλές έννοιες στα μαθηματικά (π.χ. connected set) είναι δύσκολο να δουλέψεις μαζί τους χρησιμοποιώντας απευθείας τον ορισμό. Αρκετές φορές πιο χρήσιμα από τον ορισμό είναι οι ιδιότητες και τα θεωρήματα που τον ακολουθούν.

Στην προκειμένη περίπτωση, είναι αρκετά πιο εύκολο να χρησιμοποιήσουμε τα θεωρήματα
- Path connected ==> Connected ή
- Continuous image of connected set is connected
όπως έκανε ο Durandal.


[hide]Turn that $\frac{4}{9}$ into a radical:

$\sqrt{\frac{16}{81}\cdot 3\cdot \frac{9}{2}\cdot 10\cdot \frac{90}{6}}$

$\sqrt{400}$

$20$[/hide]

[quote="Quevvy"]Could you also do this problem by simplifying each term and then multiplying?[/quote]

[hide="I guess"]$\frac{4}{9}\sqrt{3}\sqrt{\frac{9}{2}}\sqrt{10}\sqrt{\frac{90}{6}}$

$=\frac{4}{9}\sqrt{3}\frac{3}{\sqrt{2}}\sqrt{10}\sqrt{15}$

$=\frac{4}{9}\sqrt{45}\frac{3}{\sqrt{2}}\sqrt{10}$

$=\frac{4}{9}(3\sqrt{5})\frac{3}{\sqrt{2}}\sqrt{10}$

$=\frac{4}{9}(3\sqrt{5})\frac{3}{\sqrt{2}}\sqrt{2}\sqrt{5}$

$=\frac{4}{9}(3\sqrt{5})(3)\sqrt{5}$

$=4\sqrt{5}\sqrt{5}$

$=4\cdot 5$

$=20$[/hide]

I like my way better because it is my way!  :)

Also, in a competition, you way would be the fastest (and easiest!) :omighty:


Prove that the functions $ f(z) \equal{} \sin z$ and $ g(z) \equal{} \cos z$ have an essential singularity at $ \infty$

Both the functions $ f(z)\equal{}sin(z)$ and $ g(z)\equal{}cos(z)$ cannot have principal part in the expansion. Although considering the extended complex plane the "point" $ z\equal{}\infty$ forms an isolated singularity.  :blush:  :blush:


AMC
USA(J)MO
USAMO
email
AIME
AMC 10
algorithm

How are borderline people (such as myself) coping with the suspense of USAMO qualifying emails being sent out later this week? Oh, and do people know if it could possibly be earlier than Friday?

heh, well I didn't make it so I don't have to worry at all.
Good luck, potential USAMO participants.

good luck to everyone, don't get too nervous :stretcher:

And Simo, what was ur index?

Erm are they going to release a preliminary list like they did last year? Or will we have to wait until emails are sent out.

Some people think I'm borderline...but I'm not so sure (144/6)...I'm pretty sure I won't be able to make it in...but there's still some nervousness about the results...

I hope they put a preliminary list...it usually comes out a few days before the letters

I'm coping badly, knowing neither my exact AIME score(9 or 10) as well as USAMO status.

^ how early did that preliminary list come out?

Last year the preliminary list came out on the Tuesday of the same week as the USAMO qualifier emails.
This year it hasn't. Any ideas why?

maybe it will come later in the day? or tomorrow?

Just kinda... surviving throughout the day. Don't really think about it, do other stuff, etc. I'm a 129/8, and assuming floor is 7 I think I would make it, except I bubbled in a hurry near the end of the AIME and I might've misbubbled  :blush:

I'm at 150/7 (via AMC10b/AIMEII), and this will be my first year of USAMO.

Good luck everyone!!

Does anyone know what the Blue MOP cutoff score was like in the previous years?

yeah i know what you mean.. im a 124.5/8.. in a worse situation than you. if the floor is 6, i think the index would be too high for me to make it. a floor of 7, probably make it. so does anyone think the preliminary list may come out later today or tomorrow? (or if there will be one)

144/6...don't think I'll make it, but I'll be on the edge.

I'm way below borderline (124.5/7) so I guess I have no right to say anything, but still, that 0.1% chance is killing me...

I've got 150 (please donlt kill me) on AMC 10A, and (perhaps) 8 or 9 on AIME 2.

I'm not sure about my score either, though... perhaps the index will be higher this year?

Now it's up. [url]http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2009-ua/2009usamoqual.shtml[/url]

Read [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2009-ua/2009usamoqual.shtml]this[/url] for the selection info.
Here's the preliminary list (pdf).

http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2009-ua/09-Qual_list.pdf

http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2009-ua/09-Qual_list.pdf

Gah if only I could've taken AIME I.

Note that the title still says "2008 USAMO Qualifiers"... but it's actually the 2009 USAMO qualifiers :D

Grrr... no one else from Bedford made it... :(

[quote="simo14"]gah i didn't make it... why did the second index have to be 215???  :([/quote]
yes from a 202 last year wow

[quote="worthawholebean"]The culprit for that is this year's easy AMC 10.[/quote]

it was hard compared to last years.. i thought.

[quote="shelly32494"][quote="AIME15"]WHAT

A fourth grader qualified!?!?!?!?!?

Ma Michael 4 Wyatt Elementary Plano TX USA

:o

Also, a fifth grader qualified.[/quote]

ikr?! scaryy & embarasssing. hahaha. he went to the camp i went to this summmer :D
fifth grader? where?[/quote]
I thought he might qualify...

lol the 4th grader who made it was in my class in amsp...


If there were only $2$ minutes left before the proctor collects your AIME test, and you had some questions that were not answered, what would you guess on them? (or would you leave them blank?)

I would read the problem and take an educated guess.
If it asks for the length of a side, and you know about how long it is, guess around that.
If I have no clue, like on #11, I guess sumthing random that seems probably like 320.

Just guess whatever you think Zachary Abel will put for that problem. :lol: Seriously though, what hspotter said are the most commonly occuring answers. Too bad 025 didn't come up on AIME I this year :lol:


Suzy the cat has grown up to a size which enables her to climb stairs one or two at a time. How many different ways can Suzy climb a staircase consisting of ten steps?

Lets find a pattern:

1 step: 1 way (1 little step)
2 steps: 2 ways (2 little steps, or one big one)
3 steps: 3 ways (3 little steps, one big one and one little, or litte and big)
4 steps: 5 ways (4 little, 2 little one big, big little little, little big little).

etc.

This is fibbonacci, so $ 89$.

We have $ a_n\equal{}a_{n\minus{}1}\plus{}a_{n\minus{}2}$ because all paths that go up $ n$ stairs either go to the $ n\minus{}2^{th}$ step and then jump to the $ n^{th}$ step, or go to the $ n\minus{}1^{th}$ step and go to the $ n^{th}$ step.


To avoid confusion, we should reserve our posts in advance.

A good image host is here:

http://www.imageshack.us
http://www.photobucket.com

Those without photoshop can look into the GIMP, which is a fine, free alternative.

EDIT: I take that back, I suck at photoshop. Next person go please

[quote="MustItAlwaysBeMe"]So here it is.  I am somewhat doubting the Photoshop skills of the people here, but I guess it's still worth it.  Your goal is simply to make it as authentic yet funny as possible.  You are allowed to change only [b]ONE[/b] detail.  Anyone can join and play.

[img]http://upload.wikimedia.org/wikipedia/commons/1/15/Declaration_independence.jpg[/img]
---[size=59]John Trumbull's [i]Declaration of Independence[/i][/size][/quote]
Wait, how is that funny?
OH WE NEED TO EDIT IT TO MAKE IT FUNNY!
Ok.

[quote="ZhangPeijin"]Wait, how is that funny?
OH WE NEED TO EDIT IT TO MAKE IT FUNNY!
Ok.[/quote]
Exactly.

Oh yeah, and why imageshack if you could just upload here? :)

Edit:  nvm, I found the change.  Nice. ;)

Wait you talkin to me or the other guy?
I'm pretty sure its the other guy.
Anyways, I don't have Photoshop, and I don't wanna DL GIMP, and PDN isn't powerful enough so yeah.

[quote="MustItAlwaysBeMe"]

Oh yeah, and why imageshack if you could just upload here? :)

[/quote]

I'd feel bad for AoPS's servers.

Also, it'd probably be best to put images in [hide] tags.

Ahh right, [hide]tags[/hide].

And I use the GIMP...so yeah.

Actually, you are supposed to add on to the image (look at Thomas Jefferson's head in Nexmus's image), but just add on to minicon's now that you've already submitted it.  I should seriously contribute to this sometime.

I did that with a mac.  Get [url=http://www.gimp.org/][img]http://www.gimp.org/images/news-icons/wilber.png[/img] GIMP[/url]

Dude, I can't use GIMP at all.
I somehow closed my layer window, and I can't open it...

How about as pretty as possible?
[img]http://i257.photobucket.com/albums/hh202/xHassassin/DeclarationofIndependence-1.png[/img]

[quote="MustItAlwaysBeMe"]You are allowed to change only [b]ONE[/b] detail.
[/quote]

how many did you change?

1) added your name
2) threw in the words
3) included the picture

...
Let's see you do better?
Besides, rules are meant to be broken.
:|

didn't people say that you ADDED ON to the image, sooo technically ZhangPeijin's is wrong in that sense, too.. :D

Okay, I do like the part that he "cleaned up" the image, it seemed dusty before.  :D
But that totally breaks the rules, and sure it may be funny, but a game is a game.  Please follow the rules.
 :w00tb:

It's kinda like one, I mean this [i]is[/i] in the Games and Fun Factory, right?
It is basically changing one thing, only one to make you laugh.
:)

[quote="minicon"]It's kinda like one, I mean this [i]is[/i] in the Games and Fun Factory, right?
It is basically changing one thing, only one to make you belch.
:)[/quote]

What the???  :rotfl:


Find the best $k=k(n)$ such that for every natural $n$ and real $x$ we have

$(x^2+1)^n \leq k(n)(x^2+\frac{1}{2})(x^2+\frac{1}{3}).....(x^2+\frac{1}{n})(x^2+n!)$

For $x=0$ we have $1^n\leq k(n)\frac{1}{1}\frac{1}{2}\frac{1}{3}\ldots\frac{1}{n}n!=k(n)$ But for $k(n)=1$ we can prove it by Holder, or,... it's not so hard to prove it. Are my ideas all right?

It's very nice to see some easier inequalities in this forum, because in harder ones I have no chance ...


:arrow: :starwars:  :wallbash_red:  :spider:
The figure below shows the shape of Grant’s garden. The number of items planted along some of the horizontal and vertical sides is given. Each potato requires an area 6 inches by 6 inches and each bean an area 4 inches by 4 inches for each seed that is planted. If he does not intend to leave any extra space between “rows” what is the area of the garden Grant is planting, in sq feet? Express your answer as a decimal to the nearest tenth.
 :cursing:
 :furious:
 :censored:

 :ninja:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=217462&start=2200[/url]

post #2215

Why are you posting blank Word documents?


Bob wants to use 9 distinct red beads and 36 distinct blue beads to make a necklace so taht atleeast 2 blue beads are between 2 read beads. How many ways can he do this?

the number of integer solutions of $x_{1}+x_{2}+...+x_{9}= 36$ with $x_{i}\geq 2$ is $\binom{18+9-1}{9-1}= \binom{26}{8}$. the number of ways to have a set of 36 elements partitioned into 9 parts all of size at least 2, the elements in each part ordered, and then these 9 parts ordered, is then $36!\binom{26}{8}$. (since the number of ways is clearly the same as the number of ways of first arranging the 36 elements in a line, then dividing them into 9 parts - starting from the left - of size 2.) the number of necklaces of 9 such parts is then $\frac{1}{9}\cdot 36!\binom{26}{8}$, because of cyclic permutations. for each of these necklaces, the number of ways of putting 9 distinct red beads in the 9 spaces between the 9 parts - 1 red bead per space - is of course $9!$, so the answer is $9!\cdot \frac{1}{9}\cdot 36!\binom{26}{8}=8!36!\binom{26}{8}$.


I want to practice the 2000-2004 Mathcounts School competitions. If you have them, I am willing to trade you for 1999-2009 chapter, state, and national competition tests. I also have school competitions from 2005-2009. Thanks for all your help!!!!!  :)

First post! You could also post that [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189&start=580]here[/url], as most people who look there are ready for trade.


geometry
rectangle
combinatorics unsolved
combinatorics

The vertexes of the 7*7 table colored with 2 colors
 prove that there are more than 21 rectangles of the same color of vertexes


linear algebra
matrix
invariant
linear algebra unsolved

Let $ A\in M_n(F)$ be a matrix with $ n$ distinct eigenvalues in $ \bar{F}$. Prove or disprove that for any $ B\in C(A)$, $ A$ and $ B$ are simultaneously diagonalisable, where $ C(A)$ denotes the commutator of $ A$.

PS. As far as I remember a version of this problem is stated in Horn-Johnson Matrix Analysis book.

Diagonalize $ A$. The only matrices that commute with a diagonal matrix with all diagonal entries distinct are themselves diagonal.

More interestingly, this is true provided that $ A$ and $ B$ are diagonalizable, even if it they do not have distinct eigenvalues. It's reasonably easy to show that the eigenspaces of $ A$ are $ B$-invariant, from which the result follows.


trigonometry
geometry
linear algebra
matrix

The lack of talented students studying pure science is a cause of concern for many people.

Can any one explain about a career in research? I mean , when a person can get a dignified job and earn a lot with a B. Tech degree , what does a researcher get? Is it just the enjoyment in doing work he likes. Can he earn a good sum even if not as much as an engineer?

What are the opportunities for a researcher in India?

i think [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190327]this[/url] thread will answer pretty much answer all your qsns. :)
however the tough part will be to find them amogst every other argument going on in it :D

I think only the not-so-bright and not-so-aware students go for a job after IIT....
You will find loads of the brighter ones going for a Ph. D in some science stream...

Studying in IIT and going to IIM after that and then 'selling soaps' makes little sense...

Nobody has any prob if you want to be an engineer.....
You can be an innovator, do your own research in your small lab...

The prob starts when the job starts gnawing at your basic sense of science .....
when it pays you salary to work mechanically...
(makes your brain numb)

Personally,
I will hate a job which means even a slight curtailment of my freedom and it is static and not exciting...

[quote="SOUMYASHANT NAYAK"]

Personally,
I will hate a job which means even a slight curtailment of my freedom and it is static and not exciting...[/quote]

Best thing you can do is be a F1 Driver :D :rotfl: You can drive as u like n its not static and its exciting :D

F1 isn't easy--better option is go 4 karting :D  :D
u may even emulate sachin

Actually sometimes, researchers go do get jobs in private companies R&D Dept as well. However that kind of a job curtails the freedom of the person because he has to work to develop a better toothpaste  :rotfl: . If you want to do Bio-Technology or a related stream, you have immense scope for the next few years with great cash flowing in genetics field.

Many of the IT jobs that you see around you do not involve much that maths/science. RARELY you may use some mathematics (I used some basic trig to program a clock) and physics is well-left. ;)

People here are not inspired enough to take up pure sciences. The mad rush for a job is a direct consequence of the social conditions. Our students are inspired by the glamorous life of the IT people who have no qualms in flaunting what they have.

Researchers in India are an underpaid lot. :( I'm not joking. You do have good career opportunities but the pay is <<<< Engg Jobs. Usually most researchers are also professors in some colleges just to sustain their research and develop interest in students (and hire them as assistants.  :rotfl:  )

As soumya said, if you want to earn in the six/seven figures. IIT+IIM is the way to go. But after you take a job with a leading MNC or IT company or the like, your life will get mechanical. You won't have achieved anything much as in the sense of achievement. None of us talk about the million employees of MS who developed windows or those of google who are improving Orkut. It all depends on what you are inspired to do.

I for one, want to really achieve something in Robots and bring in top class robotics labs in India. There are many areas where you can pursue research in computer oriented regions.

P.S"
http://press.web.cern.ch/press/PressReleases/Releases2008/PR03.08E.html

ADD my help as well--but maybe he won't need it :rotfl:  :rotfl:  :laugh:

[quote="bruteforce"]Thanks soumya and shreyas.

@shreyas , I'll help u set up the  world class robotics institute :rotfl:  :P  :D[/quote]

I will be the villian and set up a rival group...
Also open a theme park like the one Michael Crichton wrote abt....Cybernetic or sth...
(don't exactly remember the name)
where you will have robot call-girls.... :rotfl:
robot duels....full of adventure and excitement...

[quote="SOUMYASHANT NAYAK"][quote="bruteforce"]Thanks soumya and shreyas.

@shreyas , I'll help u set up the  world class robotics institute :rotfl:  :P  :D[/quote]

I will be the villian and set up a rival group...
Also open a theme park like the one Michael Crichton wrote abt....Cybernetic or sth...
(don't exactly remember the name)
robot duels....full of adventure and excitement...[/quote]

You have small age high talent Soumya. Don't whine, I'll let you join my company as mathematical advisor and you can do all the math work involved. I'll give you complete freedom so that you'll have job satisfaction and give you 3 meals a day. Isn't that enough?

That's talking too muchhh :rotfl:  :rotfl:  :rotfl:

he will rely on CHUMMA JUJUBI!!!!!!!!!!!!! :rotfl:  :rotfl:  :rotfl:

[quote="SOUMYASHANT NAYAK"][quote]You have small age high talent Soumya. Don't whine, I'll let you join my company as mathematical advisor and you can do all the math work involved. I'll give you complete freedom so that you'll have job satisfaction and give you 3 meals a day. Isn't that enough?[/quote]

Ya i get you...
You are doing exactly the same mistake John Hammond did....

Calling a mathematician (like the chaos theorist in Jurassic Park) who is pessimistic enough to always point out why a complex system is bound to fail.....

My predictions:
1) Your robots will fail...
2) they will disobey Asimov's laws...
3) The world will be run over by them...
4) They will be so intelligent that you will commit suicide due to high level of inferiority complexion...
5) I will be 'THE ONE' and fulfill the prophecy and destroy the robot army...[/quote]

6) in the end, I'll kill you and say "Hasta La Vista Baby".
7) Soumyashants laws of Robotics will fail.
8) I'm calling you not because I want your help. Wise men say, keep your friends close, keep your rivals closer. This is the chinese reminder theorem, Lemma 2.

And did I ever tell you guys that I'm Issac Assimov's appointed successor and that I'm supposed to lead the world from metal to titanium.

P.S: I think I'm going mad because of Soumya. I pity pardesi and all the other swadesi's who had endured this demon.

[quote="Cyber-Shreyas"]
6) in the end, I'll kill you and say "Hasta La Vista Baby".
7) Soumyashants laws of Robotics will fail.
8) I'm calling you not because I want your help. Wise men say, keep your friends close, keep your rivals closer. This is the chinese reminder theorem, Lemma 2.

And did I ever tell you guys that I'm Issac Assimov's appointed successor and that I'm supposed to lead the world from metal to titanium.

P.S: I think I'm going mad because of Soumya. I pity pardesi and all the other swadesi's who had endured this demon.[/quote]

But by my 4th prediction you are already dead....

Which in contradiction to prediction no. 6
An internal contradiction in the rules of the matrix which leads to a system failure killing everyone connected to it....

9) Wise men will die...

THE END...CHUMMA JUJUBI

EDIT for Shre: Hasta La Vista Babe (not baby)

[quote="SOUMYASHANT NAYAK"]

EDIT for Shre: Hasta La Vista Babe (not [b]baby[/b])[/quote]

Dear man/woman, u have unnecessarily touched a nerve there :mad:

[quote="SOUMYASHANT NAYAK"][quote="madness"][quote="Cyber-Shreyas"] I pity pardesi and all the other swadesi's who had endured this demon.[/quote]

Who told you he was called ghatotkacha? :D :D :rotfl:[/quote]

That story is quite old (belongs to 2003 AD) propunded by Anti-crist [b]Anit[/b]

@madness
You will always remain a baby...coz its the 2200 Ad we are in now and you seem to have not realised it...[/quote]

But anit does seem to exist now :huh: Or arent you satisfied by his existance and are doing research now on his existance as the title suggests?

that's the topic of his new project :rotfl:  :rotfl:  :rotfl:


I have a great idea. People can start and exit a normal game for a "." and countdown for a "-"

People can start a message by going like this

..--..--..

and pause for 10 seconds.

People can end messages by doing this

-------------

Other then that, the letters are in "morse code"

Please tell me what you think!
It might be a bit long, but it might be the only way of communication.

[quote="dragon96"]I have some better ideas

1) Improve chat so levans will put back chat
2) Use different chatting system
3) Just start a cd and type your answers through

That would take too long[/quote]

i don't like number 3, but the rest are good

and if you're using FTW to chat, go on AIM or something
for now, i suggest pming someone if you wanna play against them (and hope they check their pms)