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I don't think that's possible ... unless of course cutting the segments was allowed.

I agree with tarquin.

PS: Tarquin, are you out of school too? No fair! :mad: I'm the only one who has to go the library at lunch and sneak on AoPS while we're at the computer lab!  :lol:

Well, if it didn't have to be ONE polygon, then it would be easy...
GRRR! Can't do a pythag Triangle either!
I'll back to this...(After school's out)

[color=indigo]I tried different ways but it didn't work. A 3-4-5 triangles too big...[/color]

Recently, a number of "bad" problems have been posted to this forum, so I understand your temptation to claim there is no solution.  This particular problem is solvable.

[quote="236factorial"]You have 12 one-unit-long straight segments. Using them, create a polygon with an area of 3 units.[/quote]

If it's a triangle, then $\sqrt{6(6-a)(6-b)(6-c)}=3$, or $(6-a)(6-b)(6-c)=\frac{3^2}{6}=\frac{3}{2}$. That's impossible if all sides are integers.

If it's a square, the sides are $\sqrt{3}$ which is also impossible.

If it's a quadrilateral, the area is $[ABCD]=3=\sqrt{6(6-a)(6-b)(6-c)(6-d)}$.

$3=\sqrt{6(6-a)(6-b)(6-c)(6-d)}$
$9=6(6-a)(6-b)(6-c)(6-d)$
$\frac{3}{2}=(6-a)(6-b)(6-c)(6-d)$

Which is impossible for any quadrilateral w/ integral side lengths.

Yes, I know this is trig, but it's the only thing I can think of  :blush: and it only works for regular polygons with $n$ sides and side lengths of $s$.

$A_{\text {regular polygon}}=\frac{ns^2}{4}\cot{\frac{180}{n}}$
$3=\frac{ns^2}{4}\cot{\frac{180}{n}}$
$12=ns^2\cot{\frac{180}{n}}$
If $n=5$, then
$12=5s^2\cot{36}$
$s \approx 1.320493115$
If $n=6$, then
$12=6s^2\cot{30}$
$s \approx 1.07$
If you keep going, you find that it isn't an integer from 7 to 180, so don't try looking for regular polygons  :lol:  :P

perhaps it is a concave polygon...

This would make it a lot simpler, no?

Oh duh I'm stupid. Just make a 3x1 rectangle and place the other four segments inside it...

Here is a picture of the polygon.
The area shaded black is the polygon.
The red lines are broken up to show individual toothpicks.

that was so wise that it scared me. Well done Xantos C. Guin

Xantos:  That's a clever solution, but it isn't actually a "polygon".

tarquin:  Your 3x1 rectangle, with the extra four segments tossed into the center doesn't qualify as a "polygon".  There may be solutions that are "close" to a 3x1 rectangle, but you need to supply some coordinates  for the vertices (or some other justification) to show it can actually be done.

There is a very nice polygonal solution.  Keep looking.

The closest I've gotten to is around 4, which is a 5-5-2 triangle.

no solution yet for me either. I've tried caving in rectangles,

[quote="Hamster1800"]well let us assume it is a triangle. By using heron's formula, we get:

$A = \sqrt{6(6-a)(6-b)(a+b-6)}$

thus:

$9 = 6(6-a)(6-b)(a+b-6)$

so

$1.5 = (6-a)(6-b)(a+b-6)$

Since the factors are all integers it cannot be a triangle.

I'm going to work on quadrilaterals next.[/quote]

look a few posts back, I already did it.

I know you did it but i thought i might reinforce it because people were still trying to do triangles :)

Well a dart is pretty much a triangle minus a triangle of the same base.  Also, adjacent sides are congruent, so the sum on one side should be 6.  Lets make the longer side 5 and the shorter side 1.  When you connect the two "tips"  we will call that base x.  Now using herons formula we can get an equation.

$\sqrt{(5+\frac{x}{2})(\frac{x}{2})(\frac{x}{2})(5-\frac{x}{2})}-\sqrt{(1+\frac{x}{2})(\frac{x}{2})(\frac{x}{2})(1-\frac{x}{2})}=3$

$\frac{x}{2}(\sqrt{25-\frac{x^2}{4}}-\sqrt{1-\frac{x^2}{4}})=3$

$\frac{x^2}{4}(25-\frac{x^2}{4}+1-\frac{x^2}{4}-2\sqrt{25-26\frac{x^2}{4}+\frac{x^4}{16}})=9$

$-2\sqrt{25-26\frac{x^2}{4}+\frac{x^4}{16}}=\frac{36}{x^2}+\frac{x^2}{2}-26$

$-2\sqrt{25-26\frac{x^2}{4}+\frac{x^4}{16}}=\frac{x^4-52x^2+72}{2x^2}$

$-4x^2\sqrt{25-26\frac{x^2}{4}+\frac{x^4}{16}}=\frac{x^4-52x^2+72}$

$16x^4(25-\frac{13x^2}{2}+\frac{x^4}{16})=x^8-104x^6+2848x^4-7488x^2+5184$

$400x^4-104x^6+x^8=x^8-104x^6+2848x^4-7488x^2+5184$

$2448x^4-7488x^2+5184=0$

By quadratic equation:

$x^2=2$

$x=\sqrt{2}$

SO it works when there is a dart with lengths 1,5,5,1 and the tips are $\sqrt{2}$ away from eachother

[quote="nat mc"]SO it works when there is a dart with lengths 1,5,5,1 and the tips are $\sqrt{2}$ away from eachother[/quote]
That's a terrific solution!  In case others don't quite see what you describe, it is a tall, skinny dart, oriented like the old Star Trek "chevron" insignia, with the point at the top.  The two long sides of the dart are of length 5 and the two short sides of the dart are of length 1.

If you wrapped a rubber band around the dart, you would have a triangle, whose sides were $5,\thinspace 5,\thinspace \sqrt{2}$

So, the base of the dart is $\sqrt{2}$.  Using the Pythagorean Theorem, the height of the dart is $\frac{7}{2}\sqrt{2}$.

The cutout turns out to be a square of side-length 1, bisected by its diagonal.  Thus the base of the dart is the diagonal of the square ($\sqrt{2}$).  And the height of the cutout is $\frac{1}{2}\sqrt{2}$.

The net area of the dart is easily confirmed to be 3.

[hide="A hint for another nice solution"]There is a solution in which every vertex of the polygon lies on a lattice point.[/hide]

[quote="nat mc"]I see the other solution.  It is a parallelogram with base 1, height three and is skewed by 4, making the long length 5. 5+1+5+1=12

Here are the lattice points: [(0,0),(1,0),(5,3),(4,3)][/quote]
Another great solution!  There's yet another solution with the vertices on the lattice points.

[hide="Hint"]The other nice solution I see is a heptagon.[/hide]

[quote="nat mc"]Are there any others?[/quote]
I'm not sure if there are any other solutions with integer coordinates.  There are no triangles.  There are several other quadrilaterals to be discovered.

[quote="chess64"]Just wondering, how do you make those amazing diagrams?[/quote]
A bit off-topic, but...

For the drawings in this thread, I used a very old and a very basic CADD program, which outputs an HPGL file.  (Note that there are other programs that will generate good (mathematically precise) line-art.)

I used Paint Shop Pro to convert from HPGL to GIF, to resize and color the drawings, and to apply ordinary text to the drawings.

I used the AoPS LaTeX engine to generate the mathematical symbols and expressions that appear on the drawings.  (For example, $\frac{7}{2}\sqrt{2}$.)  I copied and pasted from my screen's AoPS window to Paint Shop Pro.

When mathematical precision isn't required, I sometimes skip the line-art step, and do everything in Paint Shop Pro.

[quote="nat mc"][quote="rcv"]There are several other quadrilaterals to be discovered. [/quote]
Do you know of any?[/quote]
Yes.  Start with your 5-5-1-1 dart.  Cut it in half, along the line of symmetry.  Flip one half, and attach it back the unflipped half.  (Final shape is 5-1-5-1.)

How about the convex quadrilateral with side-lengths 5, 5, 1, 1.  Borrowing (and slightly tweaking) your "dart" formula:

$\sqrt{(5+\frac{x}{2})(\frac{x}{2})(\frac{x}{2})(5-\frac{x}{2})}+\sqrt{(1+\frac{x}{2})(\frac{x}{2})(\frac{x}{2})(1-\frac{x}{2})}=3$

How about the dart with side-lengths 4, 4, 2, 2.

There are plenty more.  Be creative.

You people are pretty creative. I was pretty close to that heptagon one at one time, but for some reason, my mind trailed off.

I think the simplest one was the paralleogram


Please post your answer with steps.

1. If $a,b,c \geq 0$, prove that $ab + bc + ca \geq a \sqrt {bc} + b \sqrt {ac} + c \sqrt {ab}$.

2. Prove that $x^2 + y^2 + 1 \geq xy + x + y$ for any $x$ and $y$.

I have few questions about these too.. I did get the answer after looking at the hint of the book but I don't know how you'll see the ways to solve these..Without looking at hints... ;)

[hide="My answer1"]

Remember from AM-GM that ${\frac{1}{2} (\sqrt a - \sqrt b})^2 \geq 0$, we have $\frac{1}{2} (\sqrt {ab} - \sqrt {ac})^2 \geq 0$.

Expanding it gives, $\frac{1}{2} (ab + ac - 2 a \sqrt {bc}) \geq 0$ and then:

$\frac{ab + ac}{2} \geq {a \sqrt {bc}}$.

Apply this to other two and add them and we get the answer. QED

My problem was to see the idea of using $\frac{1}{2} (\sqrt a - \sqrt b)^2 \geq 0$.. [/hide]

[hide="My answer2"]

Again.. I had to look on the hints.. Hint actually gave the answer..

Anyway.. It was make them into all left side and get:

$\frac{(x-y)^2 + (y-1)^2 + (x-1)^2}{2} \geq 0$

Expand it and simplify back into original inequality, we're done.. QED.

But my question is.. How did book come up with such clever equation to express it?[/hide]

Any tip on how to know those stuff?

[hide="1."]
This also directly follows from rearrangement of $\sqrt{ab},\sqrt{bc},$ and $\sqrt{ca}$. Whenever you see a different arrangement of the same terms and a homogenous inequality, its good to look for rearrangement because it's usually quite simple. As for the AM-GM/Trivial solution, just look for factoring into squares whenever it's easy to move things onto one side; that's really all I can say about it - if you see it enough times, you'll get it.[/hide]

[hide="2."]
Look for a factoring of the RHS; a convenient one is $(x+1)(y+1)$ after adding $1$ to both sides. So we have

$x^2+y^2+2 \ge (x+1)(y+1)$

Now look for a good AM-GM to apply, like

$\frac{(x^2+1)+(y^2+1)}{2} \ge \sqrt{(x^2+1)(y^2+1)}$

So we have $x^2+y^2+2 \ge 2\sqrt{(x^2+1)(y^2+1)} = \sqrt{(2x^2+2)(2y^2+2)}$

Notice that from the Trivial Inequality, $(x-1)^2 \ge 0 \Rightarrow 2x^2+2 \ge x^2+2x+1 = (x+1)^2$ so

$\sqrt{(2x^2+2)(2y^2+2)} \ge \sqrt{(x+1)^2(y+1)^2} = (x+1)(y+1)$ as desired.

The solution probably looks a lot more complicated than it really is to find, because after I got to the $x^2+y^2+2$ part I was pretty sure I needed to split it up in some way, so I just tried a few - once I got $(x^2+1)+(y^2+1)$ it just worked out.[/hide]

Here is how I would do the first (without, *cough*, invoking muirhead)

$a(b+c)\geq 2a\sqrt{bc}$, sum.  We know we want to do something like this because we cannot reduce the power of the $a$'s without weakening the inequality too much.

Rule of thumb: The highest power appearing in an inequality is sacred.  Weaken it (via AM-GM or whatever) and you risk going too far -  to a point where the inequality is no longer true.

As for the second.

$x^2+y^2\geq 2xy$
$x^2+1\geq 2x$
$y^2+1\geq 2y$

This is really self-explanatory.  Take care of the highest powered term first (the $xy$), then work on the rest.

[quote]muirhead[/quote]

Is that the name of the Inequality? :D

I don't know what that is though..

Yes, the first inequality may be viewed as a special case of Muirhead.

Just using $A^2+B^2+C^2\geqq AB+BC+CA$ for $A,B,C\in\mathbb{R}$.

[quote="blahblahblah"]Rule of thumb: The highest power appearing in an inequality is sacred.  Weaken it (via AM-GM or whatever) and you risk going too far -  to a point where the inequality is no longer true.
[/quote]

What do you mean by this?

Well, to illustrate, consider the inequality $x^3+y^3+z^3\geq x^2y+y^2z+z^2x$.  If you know the rearrangement inequality, this is obvious.  So lets try and prove it by AM-GM.

Directly applying AM-GM to $x^3+y^3+z^3$ gives $3xyz$.  This is too weak, because the RHS of the original inequality gives the same thing under AM-GM.  So we need to be more careful.  Instead, what we should do is use the inequalities

$x^2+x^2+y^2\geq 3x^2y$
$y^2+y^2+z^2\geq 3y^2z$
$z^2+z^2+x^2\geq 3z^2x$

and sum cyclically to get the result.  You'll soon figure out what I mean by strong terms and weak terms - the weak terms are ones that have a bunch of similar powers mixed in, whereas the strong terms contain the larger powers.  The inequality of Muirhead is essentially based on this idea of strong and weak terms, only it phrases everything in the language of majorization.


algebra
polynomial
quadratics
quadratic formula

[quote="helpmeplz"]I'm having a hard time trying to figure out how to factor this polynomial the way my book wants me to.

How can (2x+3) (x+1) + (4x+6) (x-1)= (2x+3) (3x-1)??

Please help me you do not know how much this means to me.[/quote]
Well, you could just do this:
\begin{eqnarray*}(2x+3)(x+1)+(4x+6)(x-1) &=& (2x+3)(x+1)+2(2x+3)(x-1) \\ &=& (2x+3)(x+1)+(2x+3)(2x-2) \\ &=& (2x+3)(3x-1) \end{eqnarray*}

Thanks... I get that one now but I still have a couple more that are killing me.

(x+1) (2x+6) - (x-2) (3x+9)= (2x+3) (3x-1)

Now I'm thinking 2x+6 and 3x+9 are equal as 2/6 and 3/9 are equal so maybe I should multiply the first side by 3 and the 2nd by 2.. but when I do I dont get the same answer.

(3x-1) (2x+3) +(6x-2) (x+3)

So...

2(3x-1) (2x+3) + (6x-2) (x+3)
= (6x-2) (4x+6) + (6x-2) (x+3)
= (4x+6) (x+3)
= (3x-1) (5x+9)
but the books answer is(3x-1) (4x+9)
Why??

I have a couple more but I'll let you guys help me with these first. Sorry for bothering everybodu  :(  You guys are so helpful!

[hide]\begin{eqnarray*}(x+1)(2x+6)-(x-2)(3x+9)\\ &=& 2(x+1)(x+3)-3(x-2)(x+3)\\ &=& (2x+2)(x+3)-(3x-6)(x+3)\\ &=& ((2x+2)-(3x+6))(x+3)\\ &=& \boxed{(8-x)(x+3)}\end{eqnarray*}
I think ur book was wrong...  :maybe:
also, multiplying one of the equations by two and the other by one changes the equation  :wink:  [/hide]

[quote="lotrgreengrapes7926"][quote="helpmeplz"]Thanks... how bout this one. I have to factor a perfect square polynomial.

6x^2- 5x - 6[/quote] That's not a perfect square quadratic, but this is its factorization [hide] $(2x-3)(3x+2)$ [/hide] You just need lots of practice factoring and it will become easy. :D

Also, if you're having trouble factoring a quadratic, you can always use the quadratic formula to find its factors.[/quote]

Thanks.. may I ask how you did that? By the way my friend told me, I start with 6 (x^2 -5/6x - 1) and then square b= -25/9 and then divide it by 4 so I get -25/36. Then I get

6 (x^2 -25/36) (-25/36 -1)

6 {(x- 5/6)^2 (-61/36)}
6 { x- 5/6 -61/36) (x-5/6+61/36)

Is this right so far?

[quote="helpmeplz"]Thanks.. may I ask how you did that?[/quote]
You can recognize factors with practice. Basically, you find factors of the leading coefficient (let's call them $a,b$) and the constant (called $c,d$) so that $ac+bd$ is equal to the linear coefficient. In this case, $a=2, b=3, c=2, d=-3$ (since $2\cdot2+3\cdot-3=-5$) and the factorization is $(ax+d)(bx+c)=(2x-3)(3x+2)$.
[quote="helpmeplz"] By the way my friend told me, I start with 6 (x^2 -5/6x - 1) and then square b= -25/9 and then divide it by 4 so I get -25/36. Then I get

6 (x^2 -25/36) (-25/36 -1)

6 {(x- 5/6)^2 (-61/36)}
6 { x- 5/6 -61/36) (x-5/6+61/36)

Is this right so far?[/quote]
I'm not sure what method you're using.  :maybe:

[quote="helpmeplz"]Thanks... I get that one now but I still have a couple more that are killing me.

(x+1) (2x+6) - (x-2) (3x+9)= (2x+3) (3x-1)

Now I'm thinking 2x+6 and 3x+9 are equal as 2/6 and 3/9 are equal so maybe [b]I should multiply the first side by 3 and the 2nd by 2[/b].. but when I do I dont get the same answer.
[/quote]

You can't multiply one side of an equation by 3 and the other by 2  :wink:  When solving problems like these, look for common terms: $2x+6 = 2 (x+3)$ and $3x+9 = 3(x+3)$.

You can verify the identity by noting that $P(x)=(2x+3) (x+1)+(4x+6) (x-1)$ and $Q(x)=(2x+3) (3x-1)$ are both polynomials of second degree. Now, if they got the same values at three different points, they must be the same.

Thanks everyone.. I'm gonna take a little break but I'll be back to try and understand. The problem is I didnt take math 436 last year I took 416 so Im trying to learn 436 through the internet and its really hard. But I'll keep trying and hopefully succeed thanks to all of you.

[quote="helpmeplz"]By the way my friend told me, I start with 6 (x^2 -5/6x - 1) and then square b= -25/9 and then divide it by 4 so I get -25/36. Then I get

6 (x^2 -25/36) (-25/36 -1)

6 {(x- 5/6)^2 (-61/36)}
6 { x- 5/6 -61/36) (x-5/6+61/36)

Is this right so far?[/quote]
I'm not sure what you're trying to do there. :huh:

If you're havig trouble factoring it, use the quadratic formula: $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$

$a=6,\ b=-5,\ c=-6$

$x=\frac{5\pm\sqrt{25+144}}{12}$

$x=\frac{5\pm13}{12}$

$x=\left\{-\frac{2}{3},\ \frac{3}{2}\right\}$

Or, by factoring:

$6x^{2}-5x-6=0$

$(3x+2)(2x-3)=0$

$3x+2=0;\ 2x-3=0$

$x=\left\{-\frac{2}{3},\ \frac{3}{2}\right\}$

[quote="helpmeplz"]6x^2- 5x-6= 0

(3x+2) (2x-3)

How did you find that? I hope its not just by guessing the factors of 6.. because lets say it was 5 instead of 6 you wouldnt be able to do it.[/quote]

It is the fact that it is 6 and not 5 that lend this problem to be solved. Here is an "expanded" version of the factoring:

6x^2 - 5x - 6 = 0

Take the factors of 6(-6) = -36 in a table and then find which pair adds to -5.

Clearly, (-9)(4) = -36 is working.

So, which of these pairs add up to -5? -9 + 4 = -5

6x^2 - 9x + 4x - 6
3x (2x - 3) + 2(2x -3)
(3x+2)(2x-3)

[quote="Totally Zealous"]A classroom method:

$6x^{2}-5x-6=0$
$6 \cdot-6=-36$
Find 2 numbers that multiply to $36$ and add to $-5$.
The numbers are $-9$ and $4$.
Write the equation as:
$6x^{2}-9x+4x-6=0$

Factor the $6x^{2}-9x$ and $4x-6$ seperately.
$6x^{2}-9x$ becomes $3x(2x-3)$
$4x-6$ becomes $2(2x-3)$
So it becomes $3x(2x-3)+2(2x-3)=0$
The use distributive property to make it $(3x+2)(2x-3)=0$[/quote]

Thank you for rewriting what I said before. :mad: LOL :rotfl:

Thanks I think I get it now. Are these correct?
12x^2 -20x -3x + 5
=4x (3x-5)
=(4x-1) (3x-5)

x^2 + 5x + 6
=x^2 + 2x + 3x+6
=x (x+2) 3(x+2)

How about 6x^4 -27x^3 - 15x^2 though? All 3 have exponenets..=

[quote="helpmeplz"]Thanks I think I get it now. Are these correct?
12x^2 -20x -3x + 5
=4x (3x-5)
=(4x-1) (3x-5)

x^2 + 5x + 6
=x^2 + 2x + 3x+6
=x (x+2) 3(x+2)
[/quote]

Some corrections, in itlalics, might be typos, not sure:

12x^2 -20x -3x + 5
=4x (3x-5)[i]-1(3x-5)[/i]

I think you forgot to put the second part of that step, but your answer is correct.

x^2 + 2x + 3x+6
=x (x+2)[i]+[/i]3(x+2)
=[i](x+2)(x+3)[/i]
[quote="helpmeplz"]
How about 6x^4 -27x^3 - 15x^2 though? All 3 have exponenets..=[/quote]

Solution:
[hide]
6x^4 - 27x^3 - 15x^2
x^2 (6x^2 - 27x - 15)

Now, you can factor easily:

x^2 (6x^2 - 30x + 3x -15)
x^2 (6x(x-5) + 3(x-5))
x^2(6x+3)(x-5)
3x^2(2x+1)(x-5)[/hide]

[hide][quote="helpmeplz"]I'm having a hard time trying to figure out how to factor this polynomial the way my book wants me to.

How can (2x+3) (x+1) + (4x+6) (x-1)= (2x+3) (3x-1)??

They dont want me to do 2x^2 + 2x+3x+1 + 4x^2 -4x+ 6x-6.

Please help me you do not know how much this means to me.[/quote]

wel 4x+6=2(2x+3) so then you factor out 2x+3[/hide]


algebra
polynomial
geometry proposed
geometry

When all vertex angles of a convex polygon are equal, call it equiangular. Prove that $ p > 2$ is a prime number, if and only if the lengths of all sides of  equiangular $ p$ polygon are rational numbers, it is a regular $ p$ polygon.

Hint :

Use $p^{th}$ root of unity and its minimal polynomial  $Q(x)= x^{p-1} + x^{p-2}  + ... +1$  . :lol:
Note that $Q$ is irreducible over $\mathbb Z [x] $ .

quite useful link, see page 5. lemma 2. [url]http://reflections.awesomemath.org/2006_1/2006_1_equi_polygons.pdf[/url]

The only if part is obvious, now we prove the if part.

Let the side lengths of the polygon be $a_0,a_1,\ldots,a_{p-1}$, in that order. We know that $a_0,\ldots,a_{p-1}$ are rationals. By scaling, we may also assume that they are integers. Let $w$ be the $p$-th root of unity, hence we have $a_0+a_1w+a_2w^2+\ldots+a_{p-1}w^{p-1}=0$. The two polynomials $P(x):=a_0+a_1x+a_2x^2+\ldots+a_{p-1}x^{p-1}$ and $Q(x):=x^{p-1}+x^{p-2}+\ldots+1$ share a common root $w$. So $w$ is also a root of $P(x)-a_{p-1}Q(x)=(a_{p-2}-a_{p-1})x^{p-2}+\ldots+(a_1-a_{p-1})x+(a_0-a_{p-1})$. However it is well-known that $Q(x)$ is the minimal polynomial of $w$, hence $P(x)-a_{p-1}Q(x)\equiv0$. Thus $a_i=a_{p-1}$, which implies that the polygon is regular.


How do you proof this?

$ \{\epsilon \}^* \equal{} \{\epsilon \}$

[quote="Fermat's Little Turtle"]what does the $ *$ mean ?[/quote]

that's the kleene star:
http://en.wikipedia.org/wiki/Kleene_star

[quote="JBL"]$ A^* = \bigcup_{i \geq 0} A^{\circ i}$, where $ A^{\circ i} = \underbrace{A \circ A \circ \cdots \circ A}_{i \textrm{ times}}$.  So, if you understand why $ \{\epsilon\} \circ \{\epsilon\} = \{\epsilon\}$, you're done.

(For others: $ \epsilon$ is the empty string, $ \circ$ is concatenation, and $ ^*$ is as I've defined it -- these are notations used, e.g., in Sipser's [u]Theory of Computation[/u].)

By the way, the questions you're posting would belong much better in the Computer Science forum, so if you have similar questions in the future, perhaps you could post them there.[/quote]

Ok, is it possible to move it then, so I can get faster answer?

To make this happen first I need to show 2 things, :

$ \{\epsilon\} \subset \{\epsilon\}^*$ and $ \{\epsilon\}^* \subset \{\epsilon\}$

if $ x\in\{\varepsilon\}^*$ then $ x \cup xx \cup xxx \cup xxxx \cup...............$ is equal to $ x$, as $ x$ is an empty string.

Therefore it is true, is this correct and how you proof the other end?

JBL already gave you the idea of proof. First prove that $ \{\varepsilon\}^i\equal{}\{\varepsilon\}$ by mathematical induction:

1. $ \{\varepsilon\}^0\equal{}\{\varepsilon\}$ by definition.
2. $ \{\varepsilon\}^i\equal{}\{\varepsilon\}^{i\minus{}1}\circ\{\varepsilon\}\equal{}\{\varepsilon\}\circ\{\varepsilon\}\equal{}\{\varepsilon\}$

Then by definition of * you get the result you wanted to prove:

$ \{\varepsilon\}^*\equal{}\{\varepsilon\}^0\cup\{\varepsilon\}^1\cup\{\varepsilon\}^2\cup\cdots\equal{}\{\varepsilon\}\cup\{\varepsilon\}\cup\{\varepsilon\}\cup\cdots\equal{}\{\varepsilon\}$


let $ a,b,c \in R$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. prove that:
$ \frac {a^2}{1 \plus{} 2bc} \plus{} \frac {b^2}{1 \plus{} 2ca} \plus{} \frac {c^2}{1 \plus{} 2ab} \leq \frac {3}{5}$.

sorry $ a,b,c > 0$

[quote="reza1370"]let $ a,b,c \in R$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. prove that:
$ \frac {a^2}{1 \plus{} 2bc} \plus{} \frac {b^2}{1 \plus{} 2ca} \plus{} \frac {c^2}{1 \plus{} 2ab} \leq \frac {3}{5}$.[/quote]
Try $ a \equal{} 1$ and $ b \equal{} c \equal{} 0.$
By the way, $ \frac {a^2}{1 \plus{} 2bc} \plus{} \frac {b^2}{1 \plus{} 2ca} \plus{} \frac {c^2}{1 \plus{} 2ab} \geq \frac {3}{5}$ is true.

the reverse is true.by couchy we have:
$ (1 \plus{} 2abc(a \plus{} b \plus{} c)).LHS \geq(a^2 \plus{} b^2 \plus{} c^2)^2 \equal{} 1$
$ u \equal{} a \plus{} b \plus{} c$. so $ abc \leq \frac {u^3}{27}$ so
it suffies to prove $ (1 \plus{} \frac {2u^4}{27})\leq \frac {5}{3}$ equal to $ u \leq \sqrt 3$

I think we can immediately use Jensen on $ f(x) \equal{} \frac {1}{x}$.
We get $ 1 \ge 3abc(a \plus{} b \plus{} c)$. But from the given, $ a \plus{} b \plus{} c \le \sqrt {3}$, and so $ abc \le \sqrt {3}/9$, so it is clearly true.
Of course, this seems similar to sumita's cauchy proof.


solve equation: $ 3^x\minus{}2^y\equal{}1930^z, x,y,z \in Z$

u can easily see that none of $ x,y,z$ is lesser than $ 0$.
by checking the parity we deduce that one of these equalities must hold:
$ 1\minus{}z\equal{}0, \mbox{which is a well\minus{}known case},2\minus{}x\equal{}0, \mbox{it is trivial},3\minus{}y\equal{}0, \mbox{ which is trivial too,check mod3 }$
did i miss sth?

[hide="my solution"]IGNORE THIS POST[/hide]

It is easy to check that $ x,y,z$ are non-negative.
Take mod 3, $ \minus{} ( \minus{} 1)^y \equiv 1 (mod 3)$, implying that y is odd.
Take mod 4, $ ( \minus{} 1)^x \minus{} 2 \equal{} 2^z (mod 4)$, meaning that $ 2^z \plus{} 2 \equal{} ( \minus{} 1)^x$ (mod 4), which is impossible unless $ z \equal{} 0$.

So we have to solve for $ 3^x \minus{} 2^y \equal{} 1$, a special case of Catalan's conjecture/theorem, whose only solutions are $ (1,1)$ and $ (2,3)$. So the answer is $ (x,y,z) \equal{} (1,1,0)$ or $ (2,3,0)$.

Catalan's conjecture was proved in 2002 using advanced methods, but this special case is easier

$ 3^x \minus{} 2^y \equal{} 1$

if $ x$ is odd then $ 3^x \equiv 3 \bmod{4}$ so the highest power of $ 2$ that can divide $ 3^x \minus{} 1$ is $ 2^1$ so $ 3^1 \minus{} 2^1 \equal{} 1$ is solution.
if $ x \equal{} 2z$ is even then $ 3^x \minus{} 1 \equal{} (3^z \minus{} 1)(3^z \plus{} 1) \equal{} 2^y$ so both $ 3^z \minus{}1$ and $ 3^z \plus{}1$ must be powers of two.
The only powers of two that differ by $ 2$ are $ 2^2$ and $ 2^1$ so $ 3^2 \minus{} 2^3 \equal{} 1$ is only solution.


Hi,
I have few asks about irregular tetrahedrons.
1)Is irregular tetrahedrons net also triangle divited to the 4 smaller triangles by strokes from the middles sides original bigger triangle?
2)How can i find out a size of irregular tetrahedrons height?
3)Lie within intersection of his heigh (perpendicular from the bottom plane (for example KLM) to the upper point (for example N)) and plane KLM centre of gravity "bigger triangle" in his net?

Thanks for answers.Sorry for my english.

Knows nobody anything about it?  :(
I am need know how to construct its net, but i dont know how rules holds there.

#1 holds only if we're dealing with a [url=http://mathworld.wolfram.com/Disphenoid.html]disphenoid[/url]. Otherwise we just have 4 triangles in no particular pattern.


I have Firefox 2.0, and my OS is Fedora 8. I can't insert the image for some reason. Does this have anything to do with my browser, or is it a bug?

Your browser is [b]not supported[/b]. Upgrade to Firefox 3.0 please which is the latest version of Firefox. You can view the applets but you can't add images. As I said in the announcement I won't support old browsers, except IE 6 (because IE 6 has 50% of the market share!).

In the future threads like this will be directly closed.


hi, i need help finding the oblique asymptote of the equation:

\[f\left( x \right) ={{-10x^{2}-13x+3}\over{5x-9}}\]

Here is what i did:

Using long division,

\[{{-10x^{2}-13x+3}\over{5x-9}}= 2x-{{21}\over 5}-{{{{{\rm{174}}}\over{\rm{5}}}}\over{5x-9}}\]

Let: $Q\left( x \right) =-2x-{{21}\over 5}$, $R\left( x \right) =-{{{{{\rm{174}}}\over{\rm{5}}}}\over{5x-9}}$

As $x \to \infty$ , $R\left( x \right) \to 0$ which implies As $x \to \pm \infty$, $f\left( x \right) \to Q\left( x \right)$

Therefore, $Q\left( x \right)$ must be the oblique asymptote.

but when i graph this, the graphs seem to cross. it's very obvious. what have i done wrong?

check your long division... you understand everything else, just arithmetic mistake.

Let's see if I remember this...   :P

For an oblique asymptote, it's OK if the graph and the asymptote cross. This is because the remainder can have varying sign, so on which side of the asymptote the graph is can change. The graph will still approach the asymptote, just from a different side.

(All of this is IF your arithmetic is correct; check that first, as [b]k8reindeer[/b] has suggested.)

Well considering that $\frac{-10x^{2}-13x+3}{5x-9}=-2x-6-\frac{x+15}{5x-9}$, I'd say you made a mistake somewhere.


inequalities
Divisors
number theory
IMO Shortlist

[quote="hitlEULER"]Of course $n > 1$. So let $n = \prod_{i=1}^r p_i^{a_i}$, where $r$ is a positive integer, $p_1 < p_2 < \ldots < p_r$ are pairwise distinct natural primes and $a_i \in \mathbb{Z}^+$, for any $i = 1, 2, \ldots, r$. We need $\prod_{i=1}^r (1 + a_i)^3 = 4\prod_{i=1}^r p_i^{a_i}$, and $v_2(\mbox{LHS}) = v_2(\mbox{RHS}) > 0$, where $v_2(\cdot)$ is the $2$-adic valuation of its argument.

Well, actually $3 \mid v_2(\mbox{LHS})$, so a fortiori $p_1 = 2$ and $a_1 = 3b_1 + 1$, for a certain $b_1 \in \mathbb{N}$. Moreover similarly $a_i = 3b_i$, where $b_i \in \mathbb{Z}^+$, if $i = 2, 3, \ldots, r$. If $r = 1$, this leads uniquely to $2+3b_1 = 2^{b_1+1}$ and $b_1 = 1$ or $b_1 = 2$, viz $n = 2$ or $n = 2^7$. Hence let $r > 1$. You get $(2+3b_1)\prod_{i=2}^r (1+3b_i) = 2^{b_1 + 1} \prod_{i=2}^r p_i^{b_i}$. So necessarily $p_2 \geq 5$, as $\mbox{LHS} \equiv 2 \bmod 3$.

If $b_1 = 1$, this means $5\cdot \prod_{i=2}^r (1+3b_i) = 2^2 \prod_{i=2}^r p_i^{b_i}$, leading to $p_2 = 5$, $b_2 = 1$ and $\prod_{i=3}^r (1 + 3b_i) = \prod_{i=3}^r p_i^{b_i}$, which is possible iff both the products in the left and right sides are empty, since by Bernoulli's inequality: $p^a > (1+3)^a \geq (1+3a)$, for any $a \in \mathbb{N}$ and any real $p > 4$. The same argument proves there's no further solution for $b_1 \geq 2$, because that implies $2+3b_1 \leq 2^{b_1 + 1}$, by induction. Then $d^3(n) = 4n$ iff $n = 2$, $n = 2^7$ or $n = 2^4 \cdot 5^3$.[/quote]

 Here you have used fortori 2-adic what does these words mean
Abdurashid

Well $p$-adic numbers are completion of $\mathbb Q$ with the metric $\rho(x,y)=\parallel x-y\parallel _{p}$ which if $x=\frac{a}{b}$ which $(a,b)=1$ then if $p|a$ $\parallel x\parallel _{p}=\parallel a\parallel _{p}$ if not then$\parallel x\parallel _{p}=-\parallel b\parallel _{p}$. You could see they are like infinite digit numbers $\overline{\dots a_{2}a_{1}a_{0}.b_{1}b_{2}\dots b_{k}}$

The intended solution is probably like the following. Let $f(x) = \frac{d(x)^3}{x}$. Observe that $f(x)$ is multiplicative. Also, $n$ is two times a perfect cube. Hence, $v_2(n) \equiv 1 \pmod{3}$ and $v_p(n) \equiv 0 \pmod{3}$ for odd primes $p$.

[b]Lemma: $v_3(n) = 0$[/b]

Proof: Assume otherwise, then using multiplicativity of $f(x)$, an exponent of $3$ gets introduced to the denominator, and in fact it is not possible for this to be eliminated. $\blacksquare$.

Now we can determine through tedious casework that $f(x) < 1$ if $x$ is a prime power of a prime greater than $3$. Hence, the only way for a prime beyond this threshold to work is if $f(2^k)$ makes up for it. We can determine that $f(2^7) = 4$ and $f(2) = 4$, and $f(2^{3k+1}) < 4$ for $k > 2$. For $k=1$, the only other possibility is multiplication by $5^3$. Hence the solutions are $n = 2^7$, $n = 2$, $n = 2^4 * 5^3$, which all work.

[b]Comments:[/b] For similar problems, observe [url = http://www.artofproblemsolving.com/community/c129h1087702_2015_hmic_1_multiplicative_number_theory]HMIC 2015.1[/url], [url=http://www.artofproblemsolving.com/community/c487h585526p3463268]Spring OMO 2014.19[/url], and [url=http://www.artofproblemsolving.com/community/c6h77684p445359]Canada 1999.3[/url]

I believe the problem should read as follows,

"For a positive integer $n$, let $d(n)$ be the number of all positive divisors of $n$. Find all positive integers $n$ such that $d(n)^3=4n$."

I would appreciate if a moderator of this forum or site admin could edit the OP to match it so it'll read correctly on the contests section as well :)

p-adic evaluations are not necessary for this problem; you can solve it with some pretty silly bounds

Clearly, the prime factorization of $n$ must be of the form $n = 2^{3x-2}*\prod_{i=1}^r p_i^{3e_i}$.

Now from $d(n) = \sqrt[3]{4n}$ we have $(3x-1)*\prod_{i=1}(3e_i+1) = 2^x*\prod_{i=1}^r p_i^{e_i}$

First, the LHS of this equation clearly has no factors of 3 and thus none of the $p_i$ of the RHS can be 3.  (1)

Second, $\forall p_i > 4$ we must have $p_i^{e_i} > 3e_i+1$ (easy to prove with induction)  (2)

Third, for $x > 3$ we have $2^x > 3x-1$

Thus, we must have $x \le 3$.

Case 1: $x=1$

Then, we have $2^x = 3x-1$. Thus, from (2) we conclude that there can be no other prime factors in this case. Thus we have $n = 2^{3x-2} = 2$

Case 2: $x=2$

Then, we have $2^x < 3x-1$. This makes it possible to have other prime factors. Notice that $3x-1 = 5$ thus 5 must be one of the prime factors. So we let $p_1 = 5$. Now, if $e_1 > 1$ then we have that $2^2*5^{e_1} > 5(3e_1+1)$. Thus, concluding from (2) again we see that $e_1 = 1$

When $e_1 = 1$ we have equality. Thus, by observation (2) again n cannot have any additional prime factors and this must be the only solution in this case.

The solution is $n = 2^{3x-2}*5^{3e_1} = 2^4*5^3$

Case 3: $x=3$

In this case we have $2^x = 3x-1$ and again like in case 1 we may conclude that n has no other prime factors. Thus the only solution in this case is $n = 2^{3x-2} = 2^7$

Thus, our solutions are $\boxed{2, 2^4*5^3, 2^7}$

The answer is $n=2$, $n=128$, and $n = 2000$. These are checked to work.

In general, we can let \[ n = 2^{3r+1} \cdot p_1^{3e_1} \dots p_k^{3e_k} \] where $p_i$ are distinct odd primes, and $e_i > 0$. We assume $n > 1$ throughout (as $n=1$ fails). Then $d(n)^3 = 4n$ rearranges as \[ 2^{r+1} p_1^{e_1} \dots p_k^{e_k} 	= (3r+2)(3e_1+1) \dots (3e_k+1). \] We now split into several cases. [list] 	[*]Suppose $r = 0$ or $r = 2$, 	so the equation becomes 	\[ p_1^{e_1} \dots p_k^{e_k} = (3e_1+1) \dots (3e_k+1). \] 	Then $e_i \neq 1$ for all $i$, 	since the right-hand side is odd.

But then $p_i^{e_i} \ge 3^{e_i} > 3e_i + 1$ for $e_i \ge 2$. 	Multiplying all these gives a contradiction unless $k = 0$. 	These two cases give $n = 2 \cdot 1^3 = 2$ 	and $n = 2 \cdot 4^3 = 128$ respectively.

[*]Suppose $r = 1$, so we have 	\[ 4p_1^{e_1} \dots p_k^{e_k} = 5(3e_1+1) \dots (3e_k+1). \] 	Then $5$ is among the $p_i$, so let us assume $p_1 = 5$. 	We have two cases: 	[list] 		[*]Suppose $e_1 = 1$. 		Then the equation becomes 		$p_2^{e_2} \dots p_k^{e_k} = (3e_2+1) \dots (3e_k+1)$. 		Then and as in the first case, 		we get $e_i \neq 1$ for any $i \ge 2$, 		and thus there can be no other primes. 		This gives the solution $n = 2 \cdot 10^3 = 2000$.

[*]Now suppose $e_1 \ge 2$. 		Then we multiply the estimates 		\begin{align*} 			5^{e_1} &\ge \tfrac{25}{7} (3e_1 + 1) \quad 				\text{since } e_i \ge 2\\ 			3^{e_i} &\ge \tfrac34(3e_i+1) \quad \text{if } p_i = 3 \\ 			p_i^{e_i} &\ge 3e_i +1 \qquad \text{if } p_i \ge 7. 		\end{align*} 		to get a contradiction since 		$\frac{25}{7} \cdot \frac{3}{4} > \frac54$. 	[/list]

[*]Now suppose $r \ge 3$. 	We then multiply the estimates 	\begin{align*} 		2^{r+1} &\ge \tfrac{16}{11} (3r+2) \qquad r \ge 3 \\ 		3^{e_i} &\ge \tfrac34(3e_i+1) \quad \text{if } p_i = 3 \\ 		p_i^{e_i} &\ge 3e_i+1 \qquad \text{if } p_i \ge 5. 	\end{align*} 	This gives another contradiction 	since $\frac{16}{11} \cdot \frac34 > 1$. [/list]

A little tedious casework, but otherwise decent problem... here's the sketch of my proof.
First observe that, given $n=p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}$, $p_1<p_2<...<p_k$, $p_1$ must equal $2$ and $\alpha_1=3 \beta_1 +1$ for some integer $\beta_1$ - using similar logic we get $\alpha_i =3 \beta_i$ for $2 \leq i \leq k$. Thus our equation simplifies to:
$$2^{\beta_1+1}p_2^{\beta_2}p_3^{\beta_3}...p_k^{\beta_k}=(3 \beta_1+2)(3\beta_2+1)(3 \beta_3 +1)...(3 \beta_k+1)$$

Here, a really important thing to reduce unnecessary casework is noticing that $p_2 \neq 3$ (trivial because $3$ can't divide LHS)- this will now be extremely useful because after playing around with the problem a bit, we can immediately see there'd be a heavy usage of bounding in the problem.

The pure "number theoretical" part of the problem is now over, and all that's left is a little casework... first we see that $\frac{p_i^{\beta_i}}{3 \beta_i +1} > 1$ for $p_i \geq 5$ (the proof is trivial), hence we can directly analyse when $\frac{2^{\beta_1+1}}{3 \beta_1 +2} \leq 1$ (as we need $\frac{2^{\beta_1+1}p_2^{\beta_2}p_3^{\beta_3}...p_k^{\beta_k}}{(3 \beta_1+2)(3\beta_2+1)(3 \beta_3 +1)...(3 \beta_k+1)}=1$), and from this we get $\beta_1 \leq 2$. Now plugging in $\beta_1=2$ directly gives you the only solution of its type as $2^{3 \cdot 2 +1}=2^7$, $\beta=0$ again gives an only solution as $n=2$. We have no choice but to put in $\beta_1=1$ and that'll again give an only solution (with a little more calculation) as $2^{3+1}5^3=2000$, and we're done.

[hide=Solution][quote=ISL 2000 N2]For a positive integer $n$, let $d(n)$ be the number of all positive divisors of $n$. Find all positive integers $n$ such that $d(n)^3=4n$.[/quote]
(Sorry for any typoes)
 [b][color=#000]Solution:[/color][/b] We have, $d(n)^3=4n=8k^3=d(2k^3)^3 \implies d(2k^3)=2k$ for $k \geq 1$. Note: $n=1$ is not a solution.
$$d(2k^3)=2k ~  \text{ where } ~ ~ n=2k^3$$
If $k$ is odd, then, $d(2k^3)=2d(k^3)=2k \implies d(k^3)=k$ where $k$ is odd. So, $k^3$ is perfect square, so
$$k^3=\ell^6 ~ \text{ where } ~ ~ k=\ell^2 \implies d(\ell^6)=\ell^2$$
Let $\ell=\prod_{i=1} p_i^{v_{p_i}(\ell)}$. So,
$$\ell^2=\prod_{i=1} p_i^{2e_i}=(6e_1+1)(6e_2+1)(6e_3+1) \ldots $$
Now, Since, for $x \geq 1$, $6x+1 < 3^{2x}$ $\implies$ $p_i=2$ $\implies$ $\ell=1$ $\implies$ $n=2$. Now, Suppose $k$ is even, then,
$$d(2k^3)=2k$$
Suppose, if $k$ is a power of $2$ $\implies$ $k=2^x$ for $x \geq 1$. So Trivially,
$$3x+2=2^{x+1} \implies x=2 \implies n=2^7=128$$
If $k$ is a not a power of $2$ and is even. Let $v_2(k)=x$ $\implies$ $k=2^xp$. Now, $p$ is completely odd. Let $p=\prod_{i=1} p_i ^{v_{p_i}(p)}$. Then,
$$2^{x+1} p_1^{e_1}p_2^{e_2}p_3^{e_3} \ldots =(3x+2)(3e_1+1)(3e_2+1)(3e_3+1) \ldots $$
Take $\pmod{3}$ $\implies$ $\text{ RHS } \equiv 2 \pmod{3} $. Hence, $v_3(k)=0$. If $3x+2 \le 2^{x+1}$, then using that, for $e \geq 1$, $3e+1 < 5^e$ $\implies$ $3x+2 > 2^x+1$ $\implies$ $x=1$. Also, note that: $\text{ LHS }$ $\equiv$ $0$ $\pmod{5}$ $\implies$ $p_1=5$
$$4 \times 5^{e_1} p_2^{e_2} p_3^{e_3} \ldots =5(3e_1+1)(3e_2+1)(3e_3+1) \ldots$$
If $e_1 >1$, then using, $3e+1 < 7^e$ & Already $\text{ LHS } > \text{ RHS }$, we get contradiction and since, $e_1=1$ works, so $n=2000$. Summing everything up, $\boxed{n=2,128,2000}$ $\qquad \blacksquare$
[/hide]

Let $\tau(n)$ denote $d(n)$
Then, let's say that $n=2t^3$, then we have that $\tau(2t^3)^3=8t^3$, that implies $\tau(2t^3)=2t$.Now let's say that $t=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_k^{\alpha_k}$
We encounter two cases.The first case is when $2 \mid t$, let's say that the primes $p_1,p_2,\dots,p_k$ are all ordered in the following way $p_1 < p_2 < p_3 < \cdots p_k$.Then we have that $p_1=2$, from here we have the relation:
$$(3\alpha_1+2)\prod_{i=2}^{k} (3\alpha_i +1) = 2^{\alpha_1+1}\prod_{i=1}^{k}p_i^{\alpha_i}$$
We first examine the inequality $2^{\alpha_1+1} > 3\alpha_1+2$, we see that this holds for every $\alpha_1 > 2$, so we will examine two cases.

The first case is when $\alpha_1 = 1$, then we have that:
$$5\prod_{i=2}^{k}(3\alpha_i+1) = 4\prod_{i=1}^{k}p_i^{\alpha_i}$$
We will have that one prime is $5$, and since for every prime number greater than $5$ we have the following inequality:
$$p^{a} > (3a+1)$$
This holds for any $a \in N$ we choose, now we have the following:
$$\prod_{i=2}^{k} (3\alpha_i+1)=4.5^{\alpha_2-1}\prod_{i=2}^{k} p_i^{\alpha_i}$$
$$(3\alpha_2+1)=4.5^{\alpha_2-1}$$
From here we see that $\alpha_2=1$, and from all of this we get that $n=2^4.5^3=2000$

The second case is $\alpha_1=2$, then we have that:
$$8\prod_{i=2}^{k}(3\alpha_i+1)=2^3\prod_{i=2}^{k}p_i^{\alpha_i}$$
$$\prod_{i=2}^{k}(3\alpha_i+1)=\prod_{i=2}^{k}p_i^{\alpha_i}$$
Because of the inequality $p^a>3a+1$, for every prime greater than $5$ we have that $p_2=3$, but this won't hold up since we get $2 \equiv 0 \; (mod \; 3)$.So $3 \nmid t$, we have that $n=2.2^6=128$

The second general case is $2 \nmid t$, but this is easy since of the inequality $p^a > 3a+1$, we get that any prime greater than $5$ doesn't divide $t$, and especially we get that $3 \nmid n$, so we have that $t=1$, and so we get that $n=2$

To summarize the only solutions are $n \in \{ 2,128,2000 \}$ . . . :D

The only solutions are $n = 2, 128, 2000$, which are all seen to work. Since $2n$ is a perfect cube, we may write
        \[ n = 2^{3a -2} p_1^{3e_1} p_2^{3e_2} \dots p_k^{3e_k} . \]
        where $p_1 < p_2 < \cdots < p_k$ and $a, e_1, e_2, \dots, e_k$ are positive integers.Then the $d(n)^3 = 4n$ is equivalent to
        \[ (3a - 1)^3 (3e_1+1)^3 (3e_2 + 1)^3 \dots (3e_k+1)^3 = 2^{3a} p_1^{3p_1} p_2^{3p_2} \dots p_k^{3p_k} . \]
        We split into cases based on the value of $a$.
[list]
[*]If $a=1$, then
            \[ (3e_1 + 1)^3 (3e_2+1)^3 \dots (3e_k+1)^3 = p_1^{3p_1} p_2^{3p_2} \dots p_k^{3p_k}. \]
            However, it is easily checked $p_i^{3e_i} > (e_i + 1)^3$ for all $e_i \geq 1$ whenever $p_1 \geq 3$.Hence the only solution in this case is $n = 2$.
[*] If $a = 2$, then
            \[ 5^3 (3e_1 + 1)^3 (3e_2 + 1)^3 \dots (3e_k + 1)^3 = 2^6 p_1^{3e_1} p_2^{3e_2} \dots p_k^{3e_k}. \]
            Since 3 does not divide the left hand side, but 5 does, we must have $p_1 = 5$. Clearly $n = 2^4 \cdot 5^3$ works. Now if there exist $e_i \geq 1$ with $i \geq 2$, then we may apply our work from the previous case to deduce that $d(n)^3 < 4n$. Thus $n = 16 \cdot 5^e$. Then the equation simplifies to
            \[ 125 (3e_1+1)^3 = 64 5^{3e} . \]
            The right hand side increases much faster than the left hand side, so there is only one solution $e=1$. (Increasing $e$ by 1 multiplies the left right hand side by 125, but the left hand side increases by a factor of at most 8.) Thus the only solution is $n = 2^4 \cdot5 ^3 = 2000 $
            [*]If $a = 3$, then
            \[ (3e_1 + 1)^3 (3e_2+1)^3 \dots (3e_k+1)^3 = p_1^{3p_1} p_2^{3p_2} \dots p_k^{3p_k}. \]
which collapses into the case $a = 2$. Hence the only solution here is $n = 2^{7} = 128$.
[*] If $a \geq 4$, then
            \[ (3a-1)^3 < 2^{3a} . \]
            This implies
            \[ (3e_1 + 1)^3 (3e_2 + 1)^3 \dots (3e_k + 1)^3 <  p_1^{3e_1} p_2^{3e_2} \dots p_k^{3e_k}, \]
            which we have already shown to be impossible.
[/list]
Having exhausted all possible cases, we may conclude that these are the only solutions.

Let $n=p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}$ for $p_1<p_2<...<p_k$ where $p_1=2$. As $4n$ is a perfect cube, we write $\alpha_1=3 \beta_1 -2$ and $\alpha_i =3 \beta_i$ for $2 \leq i \leq k$.
So our final equation is
$$2^{\beta_1}p_2^{\beta_2}p_3^{\beta_3}...p_k^{\beta_k}=(3 \beta_1-1)(3\beta_2+1)(3 \beta_3 +1)...(3 \beta_k+1)$$
We notice that $p_2 \neq 3$ since $3$ does not divide the RHS. Now we rearrange our equation in this format:
$$\frac{2^{\beta_1}}{(3{\beta_1}-1)}=\frac{(3{\beta_2}+1)}{p_2^{\beta_2}}\cdot\frac{(3{\beta_3}+1)}{p_3^{\beta_3}}\cdots\frac{(3{\beta_k}+1)}{p_k^{\beta_k}}$$
If $p_2=5$, we can see that $5^{\beta_2}=25^{(\beta_2)/2}=(1+24)^{(\beta_2)/2}\geq(1+12\beta_2)$ by Bernoulli's inequality. Thus the first fraction on RHS is $\leq1$. Similarly we can prove this for other primes. Hence the RHS is $\leq1$. Thus we obtain that $\beta_1={1,2,3}$. For each of the cases, we get $n=128,2,2000$ respectively [after some calculation]

[quote=orl]For a positive integer $n$, let $d(n)$ be the number of all positive divisors of $n$. Find all positive integers $n$ such that $d(n)^3=4n$.[/quote]

Let $n=\prod_{i=1}^{r} p_i^{a_i}$ then $d(n)=\tau(n)=\prod_{i=1}^{r}(a_i+1)$ now
$(\prod_{i=1}^{r}(a_i+1))^3=4*\prod_{i=1}^{r}p_i^{a_i}\implies p_i\neq 3$ when$p_k\neq 2, 3|a_k\implies a_k=3*b_k$ when$p_k=2, a_k\equiv 1\mod 3\implies a_k=3*b_k+1$
Now let for all  $p_i\neq 3,b_i\geq 1$ then $(3*b_i+1)<p_i^{3*b_i}$ and if $p_k=2$ then for all $b_k>3$ or $3*b_k+1>10$ then $(3*b_k+2)^3<2^{3b_k+1}$ so let
$n=2^{3b_1+1}*\prod_{i=2}^{r}p_i^{3*b_i}$
Clearly as mentioned above $b_1<3$ is only possible..
Case 1- $b_1=0$ we have
$2^3*\prod_{i=2}^{r}(3*b_i+1)=4*2*\prod_{i=2}^{r}p_i^{3*b_i}$ so clearly in this case $n=2$ is only solution.
Case 2- $b_1=1$ then we have
$5^3*\prod_{i=2}^{r}(3*b_i+1)=4*2^4*\prod_{i=2}^{r}p_i^{3*b_i}$ so clearly there must exist $p_k=5, a_k=3$ then
$5^3*4^3*\prod_{i=3}^{r}(3*b_i+1)=4*2^4*5^3*\prod_{i=3}^{r}p_i^{3*b_i}$ which $\prod_{i=3}^{r}(3*b_i+1)=\prod_{i=3}^{r}p_i^{3*b_i}$ for $p_i>5$ which is clear contradiction so $n$ has only 1 prime divisor $2$ . Hence only solution in this case is $n=2^4*5^3$
Case 3-) $b_i=2$ then $8^3*\prod_{i=2}^{r}(3*b_i+1)=4*2^7*\prod_{i=2}^{r}p_i^{3*b_i}$ we get
 $\prod_{i=2}^{r}(3*b_i+1)=\prod_{i=2}^{r}p_i^{3*b_i}$ for $p_i>3$ this is clear contradiction so $n$ has only 1 prime divisor $2$, Hence in this case only solution is $n=2^7$
Case 4- $b_i=3$ then $11^3*\prod_{i=2}^{r}(3*b_i+1)=4*2^{10}*\prod_{i=2}^{r}p_i^{3*b_i}$
But $5^3*\prod_{i=2}^{r}(3*b_i+1)< 4*2^4*\prod_{i=2}^{r}p_i^{3*b_i}$ so this case has no solution!!
Hence $n\in\{2, 2^4*5^3, 2^7\}$ $\blacksquare$

The answer is $\boxed{n = 2, 2^7, 2^45^3}$.

From the given equation, we can express $n$ as $$n = 2^{3\alpha + 1}p_1^{3\alpha_1}\cdots p_k^{3\alpha_k}.$$

$d(n)^3 = 4n$ turns into $$\left((3\alpha + 2)(3\alpha_1 + 1)\cdots (3\alpha_k + 1)\right)^3 = 2^{3\alpha+3} p_1^{3\alpha_1} \cdots p_k^{3\alpha_k} \qquad \qquad (1)$$or $$(3\alpha + 2)(3\alpha_1 + 1)\cdots (3\alpha_k +1) = 2^{\alpha + 1}p_1^{\alpha_1}\cdots p_k^{\alpha_k}. \qquad \qquad (2)$$

From $(1)$, note that $3 \nmid n$. In particular, $p_i \ne 3$ for any $i$.

Consider $\alpha = 0$. From $(2)$, we have $$(3\alpha_1 + 1)\cdots (3\alpha_k + 1) = p_1^{\alpha_1}\cdots p_k^{\alpha_k}.$$Dividing by the LHS gives $$1 = \frac{p_1^{\alpha_1}\cdots p_k^{\alpha_k}}{(3\alpha_1 + 1)\cdots (3\alpha_k + 1)}. \qquad \qquad (3)$$

If $k = 0$, the RHS is $1$, giving $n=2$.

If $k \geq 1$, $p_i \geq 5$, so$$\frac{p_i^{\alpha_i}}{3\alpha_i + 1} \geq \frac{5}{3(1) + 1} = \frac{5}{4}. \qquad \qquad (4)$$This is indeed the minimum value since the numerator increases faster (exponentially) than the denominator (linear) as $\alpha_i$ increases. Clearly, the RHS of $(3)$ is then greater than $1$, and thus there are no solutions for $k \geq 1$.

Now, let $\alpha = 1$ in $(2)$. This gives, after dividing, $$\frac{5}{4} =  \frac{p_1^{\alpha_1}\cdots p_k^{\alpha_k}}{(3\alpha_1 + 1)\cdots (3\alpha_k + 1)}.$$The minimum of $(4)$ only occurs at $p_i = 5, \alpha_i = 1$. Thus, the equation above is forced to be $$\frac{5}{4} = \frac{5^1}{3(1) + 1},$$giving $n = 2^45^3$.

Lastly, let $\alpha \geq 2$ in $(2)$. We find $$1 \leq \frac{2^3}{3(2) + 2}\frac{p_1^{\alpha_1}\cdots p_k^{\alpha_k}}{(3\alpha_1 + 1)\cdots (3\alpha_k + 1)} =  \frac{p_1^{\alpha_1}\cdots p_k^{\alpha_k}}{(3\alpha_1 + 1)\cdots (3\alpha_k + 1)},$$which is similar to $(3)$ with equality $k=0$, giving $n = 2^7$.

Having exhausted all cases, we are done. $\blacksquare$

We claim that $n=2,128,2000$ are the answers. These all work, so we prove that these are the only ones. First, notice that $4n$ must be a perfect cube, so let $n=2k^3$. Now the condition changes to $d(2k^3)=2k$, and we would still like to solve for all positive integer $k$.

Suppose that $x=v_2(k)$, and let $l=\frac{k}{2^x}$. Note that $l$ is odd. This means that the equation reduces to $$d(2^{3x+1}l^3)=2\cdot 2^x\cdot l \Rightarrow (3x+2)d(l^3)=2^x\cdot l\Rightarrow \frac{d(l^3)}{l}=\frac{2^{x+1}}{3x+2}.$$ Now if we let $l=p_1^{e_1}p_2^{e_2}\ldots p_q^{e_q}$, for odd primes $p_i$ and positive integer exponenets $e_i$, we get $$\prod_{i=1}^q \frac{3e_i+1}{p_i^{e_i}}=\frac{2^{x+1}}{3x+2}.$$ Notice that when $e_i>1$, the expression $\frac{3e_i+1}{p_i^{e_i}}$ is always less than one, and when $e_i=1$, $p_i=3$ is the only way to make the expression greater than one. Thus the maximum value of the LHS is $\frac{4}{3}$. If $x>2$, then $RHS>\frac{4}{3}$, so we must have $x\le 2$.

When $x=0$ or $x=2$, the RHS is equal to $1$. Then either $l=1$, which works, or we must have $p_i=3, e_i=1$, causing a term in the product to be $\frac{4}{3}$. All other combinations are less than one. The only other term that does not cause the product to go below $1$ is $\frac{4}{5}$ when $p_i=5, e_i=1$, and it doesn't work, so $l=1$ is the only solution in this case.

When $x=1$, the RHS is equal to $\frac{4}{5}$. Notice that $l=5$ only creates a term of $\frac{4}{5}$ and therefore works. As every other possible product term is less than $\frac{4}{5}$ except for $\frac{4}{3}$, we must include $1$ factor of $3$ if we are to find any other solutions. The next two largest terms that don't involve the prime $3$ are $\frac{4}{5}$ from $p_i=5, e_i=1$ and $\frac{4}{7}$ for $p_i=7, e_i=1$. However, combining these terms with $\frac{4}{3}$ gives us a product less than $\frac{4}{5}$. Thus we can only include one more term, and checking the terms shows us that this is not possible.

Therefore we have three solutions:

1. When $l=1, x=0$, we get $k=1$, implying $n=2$.
2. When $l=1, x=2$ we get $k=4$, implying $n=128$.
3. When $l=5, x=2$ we get $k=20$, implying $n=2000$.

The proof is complete.

[b][color=#f00]Remarks:[/color][/b] This is the first ISL problem I've seen that includes a nonobvious use of the year number into it.

I forgot that $n$ (and therefore $k$ and therefore $l$) cannot be a multiple of $3$, it would have made the casework a lot quicker.

To begin with, observe that $n=2k^3.$ Hence, our equation rewrites as $2k=\tau(2k^3).$

Let $k=2^a\cdot 3^b\cdot 5^c\cdot d$ and assume that $d\neq 1.$ Now, observe that for all $p\geq 7$ and $m\geq 1,$ $p^m\geq 7(3m+1)/4.$

Moreover, since $f(x)=2^{x+1}/(3x+2)$ and $g_i(x)=i^x/(3x+1), \ i\in\{3,5\}$ reach their minimums in $x=1,1$ and $0$ respectively, we have \[\frac{2k}{\tau(2k^3)}=\frac{2^{a+1}}{3a+2}\cdot \frac{3^b}{3b+1}\cdot \frac{5^c}{3c+1}\cdot \frac{d}{\tau(d^3)}\geq \frac{4}{5}\cdot \frac{3}{4}\cdot 1\cdot \frac{7}{4}=\frac{21}{20}>1\]which is an obvious contradiction. Therefore, $k$ is not divisible by any prime number greater than $5.$ Let $k=2^a\cdot 3^b\cdot 5^c$ as before.

Then, we have $2^{a+1}\cdot 3^b\cdot 5^c = (3a+2)(3b+1)(3c+1)$ so clearly, $b=0.$ Thus, $2^{a+1}\cdot 5^c=(3a+2)(5c+1).$

 Some more simple, crude, boundings will yield the solutions $n=2,128,2000.$

ISL marabot solve

Since $n=1$ fails, assume $n>1$. Let the prime factorization of $n=2^{3k+1}\cdot p_1^{3a_1}\cdot p_2^{3a_2}\cdots p_m^{3a_m}$, where $p_i$ are distinct primes greater than $2$.

The equation is $2^{k+1} \cdot p_1^{a_1}\cdot p_2^{a_2}\cdots p_m^{a_m}=(3k+2)(3a_1+1)(3a_2+1)\cdots(3a_m+1)$.

Case 1: $k=0$ or $k=2$.
Then $p_1^{a_1}\cdot p_2^{a_2}\cdots p_m^{a_m}=(3a_1+1)(3a_2+1)\cdots (3a_m+1)$.

Note that $a_i\ne 1$ because the LHS is odd. For $a_i\ge 2$, we get $p_i^{a_i}\ge 3^{a_i}>3a_i+1$. This gives a contradiction unless $m=0$. In this case, $n=2^1=\boxed{2}$ or $n=2^7=\boxed{128}$. Both work.

Case 2: $k=1$.
Then we get $4p_1^{a_1}p_2^{a_2}\cdots p_m^{a_m}=5(3a_1+1)(3a_2+1)\cdots(3a_m+1)$. Clearly, it's not divisible by $3$ and is divisible by $5$, so $p_1=5$.

Note that for $i>1$, we have $p_i^{a_i}>5^{a_i}>3a_i+1$ for all positive integers $a_i$. So we have $4\cdot 5^{a_1}\le 5(3a_1+1)$. This implies $a_1=1$ and $m=1$. So $n=16\cdot 5^3=\boxed{2000}$, which works.

Case 3: $k\ge 3$.
We get $2^{k+1}\ge \frac{16}{11}(3k+2)$. Since both sides are not divisible by $3$, we get $p_i^{a_i}>5^{a_i}>3a_i+1$ for all positive integers $a_i$. This is a contradiction as it implies LHS>RHS.

Sus

$n=\frac{d(n)^3}{4},$ so if $k=\frac{d(n)}{2}$ then $n=2k^3$ where $k$ is an integer. Let \[n=2^{3e_0+1}\cdot 3^{3e_1}\cdot 5^{3e_2}\cdot \ldots\cdot p_n^{3e_n}\]where $p_i$ denotes the $i$th odd prime, $e_i$ are non-negative integers but $e_n>0$, and $p_n$ is the largest odd prime in the prime factorization of $n$.

Now, the original equation gives \[(3e_0+2)(3e_1+1)(3e_2+1)\cdot \ldots\cdot (3e_n+1)=2^{e_0+1}\cdot 3^{e_1}\cdot 5^{e_2}\cdot\ldots\cdot p_n^{e_n}.\]
Note that for $p_i\ge 5,$ $p_i^{e_i}>3e_i+1$ for all $e_i\ge 1.$ Thus, $2^{e_0+1}3^{e_1}\le (3e_0+2)(3e_1+0).$ It is not hard to see that the only solutions of this are:

$(e_0,e_1)=(0,0),(0,1),(1,0),(1,1),(2,0),(2,1).$ For each of those cases, with simple bounding, the answer is $2,128,2000$

The answers are $n=2$ and $n=128$. (Proof is rushed)

First, note that $n=2k^3$ for some integer $k$. Then note that every exponent $e_p$ of every prime $p$ in the prime factorization of $n$ satisfies $e_p\neq 2\pmod 3$ which means that $3\nmid d(n)$ so that $3\nmid n$. Now suppose that
\[k=2^{e_2}\cdot 5^{e_5}\cdot 7^{e_7}\cdots\]
Then define $f(n)=\frac{d(n)^3}{n}$ and note that $f(2)=4$. Suppose we start from $1$ and multiply by $p^{v_p(k)}$ until we reach $k$, counting the value of $f(n)$. It is important to note that $f(n)$ can't decrease on every single multiplication. For primes $p\ge 5$ note that multiplying by any power $p^{e_p}$ will decrease the value of $f$, so any increase must come from the power of $2$. Again, simple bounding gives that any increase must occur when $e_2=0$ or $e_2=2$, with both making $f$ unchanged, meaning that we can't use any primes $p\ge 5$. Our only solutions are $n=2$ and $n=128$.

ACTUALLY, if $e_2=1$, we get that we can't have $e_p>0$ for $p\ge 7$, so a finite case check gives $e_5=1$ which implies $n=2000$ is a final solution.

We claim the answers are $n=2$, $n=128$, or $n=2000$.
Note that $n=2k^3$, so $d(2k^3)=2k$. Let $k=2^{e_1}3^{e_2}5^{e_3} \cdots p_i^{e_i}$. If there is a prime $p \ge 7$ dividing $k$, then $$\frac{4}{5} \cdot \frac{3}{4} \cdot \frac{p}{4} \le \frac{2^{e_1+1}}{3e_1+2}\frac{3^{e_2}}{3e_2+1} \cdots \frac{p_i^{e_i}}{3e_i+1} = 1,$$ a contradiction. This means $k=2^a3^b5^c$. Note that we need to have $\frac{2^{a+1}}{3a+2}\frac{3^b}{3b+1}\frac{5^b}{5b+1}=1$. Note that the minimum possible value of this expression is $\frac{3}{5}$. This means if one of the values is over $\frac{5}{3}$ we get a contradiction. We now track the values each fraction can achieve less than $\frac{5}{3}$. We have that $\frac{2^{a+1}}{3a+2}$ can achieve $1,\frac{4}{5},1, \frac{16}{11}$, $\frac{3^b}{3b+1}$ can achieve $1, \frac{3}{4}, \frac{9}{7}$, and $\frac{5^c}{3c+1}$ can achieve $1, \frac{5}{4}$. If $c=0$, then the only way to achieve reciprocating fractions is if they are both $1$, so we get $k=1$ and $k=4$. If $c=1$, the only way to get the product of $\frac{4}{5}$ is through $a=1$ and $b=0$, so $k=10$. This gives all the solutions and proves there is no more so we are done.

[hide=Sketch] Let $n= p_1^{a_1}\cdot p_2^{a_2}\cdots p_m^{a_m}$, then do mod 3 with exponents and then a bit bounding to get 2,128,2000 [/hide]

[hide=Solution]
The answers are $n = \boxed{2, 128, 2000}$. These work since $$d(2)^3 = 4\cdot 2 = 8,$$ $$d(128)^3 = 4\cdot 128 = 512,$$ and $$d(2000)^3 = 4\cdot 2000 = 8000.$$

Now, we show that $n\in \{2, 128, 2000\}$.

Since $d(n)^3 = 4n$, $4n$ is a perfect cube of a positive integer. Thus, $4n = a^3$ for some positive integer $a$. Since $2\mid 4n = a^3$, it follows that $a$ is even, so $a = 2k$ for some positive integer $k$. Then, $4n = (2k)^3 = 8k^3$, so $n = 2k^3$. Thus, $\nu_2(n)$ is $1$ modulo $3$, and $\nu_p(n)$ is $0$ modulo $3$ for $p\neq 2$ being prime.

Now, let $P$ be the set of prime divisors of such a positive integer $n$ satisfying $d(n)^3 = 4n$. Then, we have that
\begin{align*}
4 &= \frac{d(n)^3}{n} \\
  &= \frac{d\left(\prod_{p\in P}p^{\nu_p(n)}\right)^3}{\prod_{p \in P}p^{\nu_p(n)}} \\
  &= \frac{\prod_{p\in P}(\nu_p(n) + 1)^3}{\prod_{p\in P}p^{\nu_p(n)}} \\
  &= \prod_{p\in P}\frac{(\nu_p(n) + 1)^3}{p^{\nu_p(n)}}.
\end{align*}
Now, note that $3\notin P$, since $\nu_p(n)\in \{0, 1\} \pmod 3$ for all primes $p$, so $\nu_p(n)+1$ is never divisible by $3$. Thus, we have that
\begin{align*}
0 &= \nu_3(4) \\
  &= \nu_3\left(\prod_{p\in P}\frac{(\nu_p(n)+1)^3}{p^{\nu_p(n)}}\right) \\
  &= \nu_3\left(\prod_{p\in P}(\nu_p(n)+1)^3\right) - \nu_3\left(\prod_{p\in P}p^{\nu_p(n)}\right) \\
  &= 0 - \nu_3\left(\prod_{p\in P}p^{\nu_p(n)}\right) \\
  &= -\nu_3(n),
\end{align*}
so $\nu_3(n) = 0$, i.e. $3\notin P$. Now, a few useful estimates:

[b]Claim 1:[/b] Let $p\geq 5$ be a prime. Then, for any integer $k\geq 1$, $\frac{(3k+1)^3}{p^{3k}} < 1$.
[b]Proof:[/b]
Since $p\geq 5$, it suffices to show that for any integer $k\geq 1$, $\frac{(3k+1)^3}{5^{3k}} < 1$, i.e. $5^{3k} > (3k+1)^3$ for all integers $k\geq 1$. We do this by induction. For $k = 1$, we get $$125 = 5^{3\cdot 1} > (3\cdot 1 + 1)^3 = 64,$$ which is true. Now, assume that for some $k\geq 1$, $5^{3k} > (3k+1)^3$. Then, note that
\begin{align*}
5^{3(k+1)} &= 125\cdot 5^{3k} \\
           &> 125\cdot (3k+1)^3 \\
           &= 5^3(3k+1)^3 \\
           &= (15k+5)^3 \\
           &> (3k+4)^3 \\
           &= (3(k+1) + 1)^3,
\end{align*}
as desired. Thus, the inductive step is complete and so the claim is true. $\square$

[b]Claim 2:[/b] Let $k\geq 3$ be an integer. Then, $\frac{(3k+2)^3}{2^{3k+1}} < 4$.
[b]Proof:[/b]
It suffices to show that for any integer $k\geq 3$, $4\cdot 2^{3k+1} > (3k+2)^3$. Again, we use induction. For $k = 3$, we have $$4096 = 4\cdot 2^{3\cdot 3+1} > (3\cdot 3+2)^3 = 1331.$$ Now, assume that for some $k\geq 3$, $4\cdot 2^{3k+1} > (3k+2)^3$. Then, note that
\begin{align*}
4\cdot 2^{3(k+1)+1} &= 8\left(4\cdot 2^{3k+1}\right) \\
                    &> 8(3k+2)^3 \\
                    &= (6k+4)^3 \\
                    &> (3k+5)^3 \\
                    &= (3(k+1)+2)^3,
\end{align*}
as desired. Thus, the inductive step is complete, and so the claim is true. $\square$

Thus, by Claim 1 and Claim 2, it follows that if $\nu_2(n) = 3k+1$ for some nonnegative integer $k$ (it must be of this form, as discussed earlier), then $k\in \{0, 1, 2\}$, else by Claim 2 $\frac{(\nu_2(n)+1)^3}{2^{\nu_2(n)}} < 4$, and since for all primes $p\in P$ with $p > 2$ (as $p\neq 3$, so $p\geq 5$), $\frac{(\nu_p(n)+1)^3}{p^{\nu_p(n)}} < 1$ by Claim 1 ($\nu_p(n)$ for all $p\neq 2$ is divisible by $3$, as discussed earlier), it would then follow that $\prod_{p\in P}\frac{(\nu_p(n)+1)^3}{p^{\nu_p(n)}} < 4$, absurd, since in fact $\prod_{p\in P}\frac{(\nu_p(n)+1)^3}{p^{\nu_p(n)}} = 4$. Thus, $\nu_2(n) \in \{1, 4, 7\}$.

If $\nu_2(n) \in \{1, 7\}$, then note that $$\frac{(\nu_2(n) + 1)^3}{2^{\nu_2(n)}} = \frac{(1+1)^3}{2^1} = \frac{(7+1)^3}{2^7} = 4.$$ Thus, there exists no $p\in P\setminus \{2\}$, since if there did, we would have by Claim 1 that
\begin{align*}
4 &= \prod_{p\in P}\frac{(\nu_p(n)+1)^3}{p^{\nu_p(n)}} \\ &= 4\left(\prod_{p\in P\setminus \{2\}}\frac{(\nu_p(n)+1)^3}{p^{\nu_p(n)}}\right) \\ &< 4,
\end{align*}
absurd. Thus, in these cases, we have $n = 2^{\nu_2(n)}$, which is $2$ or $128$ based on whether $\nu_2(n) = 1$ or $\nu_2(n) = 7$. Thus, we have $n\in \{2, 128, 2000\}$.

Else, if $\nu_2(n) = 4$, we have $$\frac{(\nu_2(n)+1)^3}{2^{\nu_2(n)}} = \frac{(4+1)^3}{2^4} = \frac{125}{16}.$$ Thus, we must have $5\in P$, since we need $\nu_5\left(\prod_{p\in P}\frac{(\nu_p(n)+1)^3}{p^{\nu_p(n)}}\right) = 0$, so since $\nu_5\left(\frac{125}{16}\right) = 3$, we must have $\nu_5\left(\prod_{p\in P\setminus \{2\}}\frac{(\nu_p(n)+1)^3}{p^{\nu_p(n)}}\right) = -3$. However, if $p\neq 5$, then $\nu_5\left(\frac{(\nu_p(n)+1)^3}{p^{\nu_p(n)}}\right) \geq 0$, thus we need $5\in P$.

[b]Claim 3:[/b] If $k\geq 2$, then $\frac{(3k+1)^3}{5^{3k}} < \frac{64}{125}$.
[b]Proof:[/b]It suffices to show that for all $k\geq 2$, $5^{3k} > \frac{125}{64}(3k+1)^3$. For $k = 2$, we have $125^2 > \frac{125}{64}\cdot 7^3$, since $125^2\cdot \frac{64}{125} = 20^3 > 7^3$. Now, assume that for some $k\geq 2$, $5^{3k} > \frac{125}{64}(3k+1)^3$. Then,
\begin{align*}
5^{3(k+1)} &= 125\cdot 5^{3k} \\
           &> 125\cdot \left(\frac{125}{64}(3k+1)^3\right) \\
           &= \frac{125}{64}(15k+5)^3 \\
           &> \frac{125}{64}(3k+4)^3 \\
           &= \frac{125}{64}(3(k+1)+1)^3,
\end{align*}
as desired, so the inductive step is complete and so the claim is true. $\square$

Now, note that $\nu_5(n) = 3\ell$ for some positive integer $\ell$ (positive since $5\in P$).

If $\ell = 1$, then we have $P = \{2, 5\}$, since if $P\neq \{2, 5\}$, then we have by Claim 1 that
\begin{align*}
4 &= \prod_{p\in P}\frac{(\nu_p(n)+1)^3}{p^{\nu_p(n)}} \\
  &= \frac{(\nu_2(n)+1)^3}{2^{\nu_2(n)}}\cdot \frac{(\nu_5(n)+1)^3}{5^{\nu_5(n)}} \cdot \prod_{p\in P\setminus \{2, 5\}}\frac{(\nu_p(n)+1)^3}{p^{\nu_p(n)}} \\
  &< \frac{(\nu_2(n)+1)^3}{2^{\nu_2(n)}}\cdot \frac{(\nu_5(n)+1)^3}{5^{\nu_5(n)}} \\
  &= \frac{(4+1)^3}{2^4}\cdot \frac{(3+1)^3}{5^3} \\
  &= 4,
\end{align*}
a contradiction. Thus, $P = \{2, 5\}$, so $n = 2^{\nu_2(n)}5^{\nu_5(n)} = 2^45^3 = 2000 \in \{2, 128, 2000\}$, as desired.

Else, we have $\ell \geq 2$. Then, by Claim $3$, we have
\begin{align*}
4 &= \prod_{p\in P}\frac{(\nu_p(n)+1)^3}{p^{\nu_p(n)}} \\
  &= \frac{(\nu_2(n)+1)^3}{2^{\nu_2(n)}}\cdot \frac{(\nu_5(n)+1)^3}{5^{\nu_5(n)}} \cdot \prod_{p\in P\setminus \{2, 5\}}\frac{(\nu_p(n)+1)^3}{p^{\nu_p(n)}} \\
  &\leq \frac{(\nu_2(n)+1)^3}{2^{\nu_2(n)}}\cdot \frac{(\nu_5(n)+1)^3}{5^{\nu_5(n)}} \\
  &< \frac{(4+1)^3}{2^4}\cdot \frac{64}{125} \\
  &= 4,
\end{align*}
where we set $$\prod_{p\in P\setminus \{2, 5\}}\frac{(\nu_p(n)+1)^3}{p^{\nu_p(n)}} = 1$$ if $P = \{2, 5\}$.

Thus, in all cases, $n\in \{2, 128, 2000\}$, and we have shown that all three work, so we're done. $\blacksquare$

This problem can be cracked by making a series of observations.

Consider $d(n)^3 = 4n$. This means that at a structural level, $n = 2j^3$ for some positive integer $j$. Thus we know that $d(n)^3 = 4n = 8j^3 \implies d(2j^3) = 8j^3 \implies d(2j^3) = 2j$.

Now we can arbitrarily think of $j = 2^{e_2}3^{e_3}5^{e_5}7^{e_7} \dots$. Then note that $d(2j^3) = d( 2^{3e_2+1}3^{3e_3}5^{3e_5}7^{3e_7} \dots) = (3e_2 + 2)(3e_3 + 1)(3e_5 + 1) \dots \equiv 2 \pmod{3}$. So if $d(2j^3) \equiv 2 \pmod{3}$ for all $j$ it must be true that $2j \equiv 2 \pmod{3} \implies j \equiv 1 \pmod{3}$.

$j = 1$ obeys $d(2) = 2$. Now consider odd $j \geq 7$. Its prime factorization assumes the form $5^{e_5}7^{e_7}11^{e_{11}} \dots$. If we look at $d(2j^3) = 2(3e_5 + 1)(3e_7 + 1)(3e_{11}+1)\dots$. This needs to equal $2j = 2 \times 5^{e_5}7^{e_7}11^{e_{11}} \dots$ implying that $(3e_5 + 1)(3e_7 + 1)(3e_{11}+1)\dots = 5^{e_5}7^{e_7}11^{e_{11}} \dots$. But note that $p^m > 3m + 1$ for $m \geq 1$ and $p \geq 5$.  So this would not be possible.

Thus it must hold that if $j > 1$, it is even. Now let us begin considering $v_2(j)$. If $v_2(j) = 1$, then we now that $j = 2 \times 3^{e_3}5^{e_5}7^{e_7} \dots$. Then $d(2j^3) = 5(3e_5 + 1)(3e_7 + 1) \dots = 2j = 4 \times 5^{e_5}7^{e_7} \dots$. This only holds if $e_5 = 1$.

If $v_2(j) = 2$, we see that the only valid option is no other primes dividing $j$ since $d(2j^3) = 8 \times (3e_5 + 1)(3e_7 + 1)(3e_{11} + 1) \dots = 2j = 8 \times 5^{e_5}7^{e_7}11^{e_{11}} $. Since $p^m > 3m + 1$ for $m \geq 1$ and $p \geq 5$, it must hold that $e_5 = e_7 = e_{11} = 0$.

If $v_2(j) = e_2 \geq 3$, then $3e_2 + 2  < 2^{e_2 + 1}$ not allowing equality.

So $j = 1, 4, 10 \implies n = 2j^3 \implies n = 2, 128, 2000$


Do there exist positive integers $m$ and $n$ such that:
\[ 3n^2+3n+7=m^3 \]

[hide]m is odd since if n is even, even+even+odd and if n is odd, odd+odd+odd.  Now let's look at this mod 3.  Cubing the term on the right yields 0 if you put in 0, 1 if you put in 1, and 2 if you put in 2.  The left hand side will have 0+0+7 or equaling to 1 mod 3.  In other words, 3x+1 are the form our numbers come in.  We substitute 3x+1 for m where x is even.  (3x+1)^3=27x^3+27x^2+9x+1=3n^2+3n+7.  We simplify to get 9x^3+9x^2+3x-2=n^2+n.  Now let's consider this in mod 3 again.  For the terms on the right, we get 0+0+0-2 or 1 mod 3.  The left hand side is 0 when 0 is inserted, 2 when 1 is inserted, and 0 where 2 is inserted.  In other words, it never equals one mod 3 so substituting 3x+1 for m causes an answer to be nonexistant but that's the only form the answers come in from the first part of the proof.  In other words, there's a contradiction so there is no positive integers m and n.[/hide]
Did I do this right?  [hide]Here's why I don't think it's right.  If you do the mod 3 thing immediately after we substitute 3x+1 in, we don't have a problem.  But when we analyze with mods after simplification, we get the contrdiction. [/hide]

It's more or less correct, but there is a much quicker way to do this:

[hide]mod 9.[/hide]

Yes , Arne , mod 9 works very fast

[hide]

We can change it into $(n+1)^3-n^3+6=m^3$

Notice that the residue of a cube mod 9 can only be $\pm 1,0$ . And for difference of consecutive cube , the only possiblity is $(n+1)^3-n^3\equiv 1,-2\mod 9$ . So LHS$\equiv 4,7\mod 9$ which obviously shows no solution for $m^3$ .

[/hide]


1.  How many mLs of hydrogen gas can be produced in the reaction of 3.0 M Hydrochloric acid and 2.50 g grams of Zinc?

2.  How many mLs of 3.0 M HCl is needed to dissolve 2.50 grams of Zinc?

3.  How many molecules of Zinc Chloride are produced from 2.50 grams of Zinc?

4.  If you reacted 6.00 grams of Copper (II) Sulfate and 2.50 grams of Zinc, how much Zinc Sulfate would be produced and how much of which reactant would be leftover?

5.  If you started with 2.43 grams of copper, what is the molarity of copper (II) nitrate solution found after you added 20mLs of 15.9 M nitric acid and 100 mLs distilled water?

 :D

For the first three exercises, start by writting the balanced chemical equation:

$ 2HCl \plus{} Zn \longrightarrow ZnCl_2 \plus{} H_2$

Now:

- How many moles are there in 2.50 g of zinc?
- What is chemical quantity of hydrogen that can then be produced?
- Assuming NTP conditions, what is the corresponding volume?

What is NTP, Carcul?

I've only ever heard of STP, standard temperature and pressure.

EDIT:  Oh, is it normal temperature and pressure?  What is that then?  Room conditions, $ 298 \text{K}$ and $ 1 \text{atm}$?


I know this inequality is correct, but I would like to see the solution:

$ \sqrt{\frac{x}{y\plus{}z}}\plus{}\sqrt{\frac{y}{x\plus{}z}}\plus{}\sqrt{\frac{z}{x\plus{}y}}\geq \frac{3}{\sqrt{2}}.$

The best bound is $ 2$. Check what happens when one of the numbers is zero. Also, here is a strenghtening: http://www.mathlinks.ro/Forum/viewtopic.php?t=149855


ARML Support geometry AMC AIME USA(J)MO USAMO

I heard a lot of complaints at the Las Vegas site, so I decided to start this thread to see how many people were dissatisfied with UNLV. I would like to see people's comments, suggestions, and experiences.

My experiences: A ten hour bus ride to UNLV (compared with about half an hour to San Jose State), REALLY HOT WEATHER (triple digits every day), individual rounds in a room that was HOT and HUMID (felt hotter than outside).

Apparently the site coordinator even passed out from heat exhaustion.

This is a petition of sorts, post if you want better facilities for ARML to hold its Western Division site at (maybe UNLV, San Jose State, maybe elsewhere).

I'm not sure where you're trying to go with this... yes, there were some things that could've gone better, no doubt many of them because of the heat, but it wasn't bad to the extent of needing drastic change. I myself am willing to go to Las Vegas ARML for another year (my last) as long as they find an air-conditioned room for us to test in. Even if they don't, repeating this year's individual results would be fine by me :P.

I agree with paladin8. Nothing wrong with Las Vegas itself. Plenty to do. The housing complex was pretty good -- friendly staff, games, TV. I guess most teams were there long enough to utilize it like we did. On the other hand I think the testing room absolutely needs to be changed if it's going to be at UNLV again. There's no way I should be dripping sweat onto my paper and dealing with a headache from the heat.

I still liked ARML at UNLV, and most people who are willing to travel so far for a math competition probably won't dislike it just because of a few hours in absurd heat (and the resulting problems).

I support the petition to have ARML somewhere else, preferrably back on the coast where it's not as hot outside, too.

The temperature was certainly a problem and the goal is to find much better accomodations for testing next year.  We can also hope that it won't be unseasonably warm this time next year.

I'm not sure the weather was unseasonably warm, by the way --
ARML was at UNLV from 1995-2000, before switching to San Jose State. it was over 100 degrees at least three of those six years, the other years it was in the 80s and 90s.

They never did have a good place for the individual round, though the old location wasn't as bad as this year's, judging from everyone's comments. But I would definitely agree that this ought to be a solvable problem. Heck, maybe ARML can try getting some help from the Las Vegas Convention Center. A city that can host Comdex can find a place for the individual round.

As for the rest, I think Las Vegas may well be the best available site -- it may be a longer drive for the SFBA teams, but it's a  shorter trip for many other teams, and also a cheaper flight.  The fact remains that we used to have 20-24 teams at UNLV, then it dwindled to about a dozen at San Jose State and now it's back up again.

I thought it was fine.

The only thing i dont like is that in other arml sites they read off the question while you look at it (i think).

I won't belabor the things that have to be fixed - everyone knows what they are by now.

For us, the transportation difference between San Jose and Las Vegas is only an hour or two of bus ride, but Las Vegas is a little closer and a little easier from a traffic point of view.

I missed the Friday evening cookout we had at San Jose which was a place for people from different teams to mingle. Is there any way some kind of event (preferably involving food) could be staged on Friday afternoon in a large enough space? The problem with that, of course, is that nowhere outside is particularly attractive because of the weather.  (As Ted pointed out, the outdoor temperature, while probably higher than the average for the date, was nothing at all unusual for Las Vegas.)

We were over in the Dayton complex of dorms. We took over one of the lounges, on the third floor, which was good for us as a team but reduced our contact with the other teams.

If a proper room can be found to stage the Saturday activities (individual round, relays, frivolous contests, awards ceremony), then I would vote to stay in Las Vegas.

Someone has mentioned the fact that the individual questions are read aloud before time starts at some sites and not others.  This should probably be standardized for fairness's sake.

I didn't think Vegas was that bad, except for the difficulty involved in playing frisbee, and the obvious air-conditioning problem that should be specific to this year.  I actually enjoyed the bus ride and the opportunity to socialize that it provided.  I have one suggestion that's only been touched on thus far: Teams were split between two dorms, Dayton and Tonopah.  Dayton worked out OK; the lounges are really, really nice and provided a perfect gathering/meeting/lunch/game area for us.  However, we had virutually zero contact with other teams, even the ones that were in Dayton, as there's no central lounge area.  I think that if possible, all teams should be housed in the same dorm next year, and that that dorm should include a central gathering area downstairs.  The Iowa site does this really well; the lobby of Burge Hall has tons of space for people to hang out, play games, etc.  IMHO, that works a lot better.

We'll find a better room for the individual contest next year, or I'll personally try to find another site for the contest.   We won't be in that room again.  I should have pulled the plug on it when I saw it.  (BTW, it wasn't the heat that got me - it was food poisoning or a stomach virus.)

I'll also be looking into reinstating something on Friday evenings.  I agree that there wasn't nearly enough interaction among the teams.

The weather was unseasonably warm, but not so much so that we can count on it not happening again...

We'll definitely standardize the way the Individual Round is run, too.  Keep the suggestions coming - we'll see what we can do to improve it.  This was our first year at a new site for most of us, so next year is bound to be much better.  (I'm just begging the karma gods for even worse weather, aren't I?)

[quote]I'm just begging the karma gods for even worse weather, aren't I?[/quote]

Oh yes you are. Just don't beg the food gods for more poisoning. ^^;

But anyways... I half agree with everything. UNLV was bad because it was hot, but I'd be okay with it if the individual round was in a different place. I overheard someone talking about how it might've been in a different place, but that it cost money to rent the room (or something along those lines). I don't think the actual bus ride was terrible, but it is much easier to drive to SJ than to drive all the way down to UNLV.

The thing that really bugs me is the lack of interaction between teams. Yah, the study room in Dayton Complex made a nice meeting place, but it didn't encourage getting to know other people? Not much of that. Or at least, not for me. I think I only talked to one or two people outside of SFBA. I mean, it's nice and all to have more teams at a site, but it's sort of pointless if you don't meet anyone...

I wasn't at Las Vegas, but on another thread, I saw someone suggest Salt Lake City, Utah.  I think that or Park City would be a good idea, climatewise, anyhow.  Several years ago, I was at Park City because my mom had a math conference there.  It felt fairly temperate, as I remember, and dry, although (according to my mom) not nearly as dry as Las Vegas.

By the way, I think that extreme temperatures could make it unfair for the competitors at Las Vegas, as at Penn State we were quite comfortable and therefore were probably overall closer to our respective full potentials.

Even though this would never happen, Seattle has very nice weather throughout the entire year and wouldn't give an unfair advantage to those teams who actually have experienced temperatures above 100 degrees. I really think that the heat on Saturday was the largest problem, particularly for most of us Washingtonians where the record temperature in Seattle is less than it was on either day of the competition. I also agree that more interaction amongst the teams would be nice, since the extent of my interaction with any other team was pretty much talking to some random people in the elevator.

I did, however, appreciate the bathrooms. Much better than at San Jose. And the dorms were pretty nice in general.

Although I doubt that anyone would take Boy Soprano II's suggestion seriously, I'll reply anyway.

There were 22 teams at Las Vegas, not counting the alternate team. What would the transportation effect be on those 22 teams of moving to Utah instead of California?

Much better: 1 - Utah.

Slightly better: 2 - Colorado, New Mexico. Colorado could drive. New Mexico is still quite a ways away - I don't know that it would make much difference to them.

Slightly worse: 9 - Washington (2), Alabama, Taiwan (4), Philippines (2). These are teams that would have to fly in any case. Remember that one of the biggest attractions of Las Vegas is its accessibilty to air travel. Even so, the Taiwan teams flew into Los Angeles and took chartered buses from there to Las Vegas. If Park City rather than Salt Lake City (and notice that Park City is not a college town - ARML prefers college campuses), then you've got the additional burden of getting 20 or 40 or however many people there from the Salt Lake airport.

Much worse: 10 - SFBA (3), Northern California, Southern California (3), San Diego (2), Nevada. The Nevada team lives there; the California teams arrived on buses. That ranges from a 4-5 hour trip for us to an 8-10 hour trip for SFBA. Moving to Utah would dramatically increase the costs and dramatically decrease the participation for the California teams.

Just from the geography/transportation point of view, Las Vegas is probably the best location. (The second best would be somewhere in the Los Angeles metropolitan area.) The transportation advantages of Las Vegas over San Jose probably had a lot to do the substantial increase in the number of teams at the western site this year. Utah would be a worse location than San Jose.

Alabama could come to Penn State—I hear they used to do that.  But aside from that, I'll trust your judgment—my knowledge of western U.S. geography is rather poor, especially with regard to distance.

For Washington, flying to Salt Lake City might actually be easier and cheaper than Las Vegas since Delta has direct flights from Seattle to Salt Lake City, and I'm sure that my fellow teammates would agree when I say that we would most definitely appreciate any site that doesn't equal death if there isn't air conditioning. But then again, if they're definitely going to change the venue for the Saturday events, I think Las Vegas is fine.

[quote="Uh Oh"]Even though this would never happen, Seattle has very nice weather throughout the entire year[/quote]
One big reason this won't happen: the University of Washington can't possibly host. This year, ARML coincided with the last day of classes and the first day of finals.

About Salt Lake City:

I think I am the person who brought this up in the other thread, so I'll post some more things about it here.

First off, I live in the suburbs of Salt Lake City, so now you know where I'm coming from.

When I brought it up in the other thread, I doubted it would happen, but here are some reasons why I thought it might be good other than it being where I live.

1.  Salt Lake Intl. Airport is a Delta hub.  This would mean that it would be cheaper airfares for people going on Delta Airlines.

2.  Cooler than Las Vegas.  From my experiences there this year, I do not want ot have to go there again.

Someone suggested Park City as a possible location.  As has been said, it does not have a major college campus.  It also is about an hour away from the airport (rough estimate), making it difficult for teams to get to it.  It would be a nice place because of all the attractions (main olympic city...), but I don't think it could be there.

The only location in Salt Lake City that could possibly work is the University of Utah.  I have been to their student housing before, and I remember the housing to be nicer than we had at UNLV.  However, I don't know of any building in particlar that we could have the contest in.  Of course I haven't been to very many buildings on campus.

So there's what Salt Lake City would be.  I don't necessarily think it would be the best location, but I don't want to go to Las Vegas again.

[quote="coloradoarml"]Except for the breakdown of the air conditioning during Saturday's competition rounds, I think UNLV was a good choice for the western ARML site.  Geographically, Las Vegas is the best central location for the western states and the least expensive to fly to.  Yes, it was hot outside but the dorms, dining hall and competition rooms were all close together, just a few minutes walk apart.  Our Dayton dorm rooms were very pleasant and comfortable.

For next year, I also would like to see more interaction between teams, though I would prefer Saturday evening activities (including a math talk, perhaps?) to Friday.[/quote]

Would your team stay for Saturday activities if they existed (and pay the requisite dorm/food cost)?  And what other activities might you suggest in addition to some math?

[quote="joml88"][quote="coloradoarml"]For next year, I also would like to see more interaction between teams, though I would prefer Saturday evening activities (including a math talk, perhaps?) to Friday.[/quote]

Many teams leave right after the award ceremony is over on Saturday (that's how it usually is at Penn State, at least).  This seems to be the major reason for why the socializing and math talk take place on Friday at Penn State.[/quote]

Friday was a busy day for us.  We left Denver in the morning, arrived in LV mid-afternoon, had a little time to settle into our dorms, ate a quick dinner, then headed straight for the Power and Team Rounds, which ended around 8:30 pm (9:30 mountain time).

We would have to arrive on Thursday if social activities were scheduled for Friday afternoon.  I'd prefer to schedule the socializing for Friday night after the competition rounds or on Saturday.

[quote="rrusczyk"]Would your team stay for Saturday activities if they existed (and pay the requisite dorm/food cost)?  And what other activities might you suggest in addition to some math?[/quote]

Yes, I think we would stay Saturday night.  This year we left immediately after the awards ceremony and didn't get a chance to wind down or chat with other teams.

As for other activities, if a UNLV song contest would draw enough participants, I'd be in favor of that.  Otherwise, a talent show?

[quote="coloradoarml"]
As for other activities, if a UNLV song contest would draw enough participants, I'd be in favor of that.  Otherwise, a talent show?[/quote]

Talent show at West Coast ARML = we all sit around and watch Macky do cool stuff.  I'm all for that :)

Oh, right.  I forgot that you guys do part of the competition on Friday  :oops:   That makes things different.

This is a plea to the guys at UNLV to have a couple spare fans around just in case the air conditioning dies again!  I thought I was going to have a heat stroke last year  :blush:  :blush: During the individual round, I spent more time fanning myself and drinking water than actually solving the problems :blush:  :blush:  :(

[quote="Phoenix257"]During the individual round, I spent more time fanning myself and drinking water than actually solving the problems :blush:  :blush:  :([/quote]

...that reminds me. In the individual round in 1998, I went into the last problem pair with a perfect 6/6. Early in the round, I started getting a [i]massive[/i] nosebleed. I mean, it was [i]fierce[/i]. It must have been a stress-related thing, since I never get nosebleeds. I had to flag down the proctor for our aisle and beg him to find some kleenex  :|

I ended up getting both questions wrong on that pair, although I'm pretty sure it was simply because they were too hard (and not because of my fire hydrant of a nose  :blush: ). Hopefully the heat won't create any stories this year at UNLV as sad / gross as mine, heh.

Just FYI, the individuals are moving to a new place this year.  According to the material on the ARML site, they're in an auditorium on the second floor of the student union.  It's located sort of "behind" Beam Hall, where some of the teams did team and power rounds, and where power rounds were graded.

Ironically, the Iowa individuals are also in an auditorium (or ballroom, minor detail) on the second floor of the student union.

I don't think I've ever wanted air conditioning as badly as I did during that test

I think it would also be very nice if we could write on a desk or table during the individual round - would that be possible?

Well, if it's an auditorium or a ballroom, that's unlikely. Back in San Jose, all we had were clipboards, so that's no real difference. Looking at the rotating pictures that show on the home page at http://www.arml.com, I see the individual or relay rounds in progress at three different locations. One of them is the auditorium in San Jose - clipboards, no chair arms. It doesn't look like there are desks or chair arms at the other two locations, either.

The Penn State Individuals were in the basketball arena last year, so definitely no desks/chair arms (most people don't do math while watching basketball . . .).  I didn't think it was that big of a problem, though.  ARML Individuals are so fast that organization isn't an issue (the AIME or USAMO would be a different story, of course!)

[quote="Klebian"][quote="Kent Merryfield"]... Looking at the rotating pictures that show on the home page at http://www.arml.com...[/quote]

Wait where is this?[/quote]

I think he means [url]http://www.arml.com./about.htm[/url] at the bottom of the page.


suppose $ f(x)\equal{}\minus{}x^2\plus{}bx\plus{}c$ and $ f(x) \leq 2$ for all $ x \in R$.

if the function has two real roots $ x_1,x_2$ , find the maximum value of $ x_1\minus{}x_2$.

$ f(x) \equal{} \minus{} (x \minus{} x_1)(x \minus{} x_2)$ then $ c \equal{} \minus{} x_1x_2$ and $ b \equal{} x_1 \plus{} x_2$

and $ (x_1 \minus{} x_2)^2 \equal{} d^2 \equal{} b^2 \plus{} 4c$

On the other hand $ f(x) \leq 2$ for all $ x \in R \iff f(\frac {x_1 \plus{} x_2}{2}) \leq 2$

$ \minus{} (\frac {b}{2})^2 \plus{} b\frac {b}{2} \plus{} c \leq 2 \iff \minus{} \frac {b^2}{4} \plus{} \frac {b^2}{2} \plus{} c \leq 2 \iff \frac {b^2}{4} \plus{} c \leq 2 \iff b^2 \plus{} 4c \leq 8 \iff d^2 \leq 8 \iff d \leq 2 \sqrt {2}$

One example for $ \boxed{d \equal{} 2 \sqrt {2}}$ is $ x_1 \equal{} 0$ and $ x_2 \equal{} 2 \sqrt {2}$ for $ \minus{} x^2 \plus{} 2 \sqrt {2}x \equal{} \minus{} x(x \minus{} 2 \sqrt {2}) \equal{} 2 \minus{} (x \minus{} \sqrt {2})^2$

it was an easy problem,i could solve it easily...


inequalities
geometry
circumcircle
inequalities proposed

$\triangle ABC\Longrightarrow \frac{1}{b+c} +\frac{1}{c+a} +\frac{1}{a+b} \le \frac{3(R+r)}{4S}.$

[b]First solution: [/b]

$ LHS \equal{} \frac {3(ab \plus{} bc \plus{} ca) \plus{} a^2 \plus{} b^2 \plus{} c^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$ .

Now, using the well-known ineq. $ (a \plus{} b)(b \plus{} c)(c \plus{} a)\ge 8abc$ obtain:

$ LHS\le\frac {3(ab \plus{} bc \plus{} ca) \plus{} a^2 \plus{} b^2 \plus{} c^2}{8abc} \equal{} \frac {5s^2 \plus{} r^2 \plus{} 4Rr}{32RS}$, where $ S$ is the area of triangle, $ s$ is the semiperimeter and $ r, R$ are the radius of incircle and circumcircle respectively.

So we'll only have to prove that: $ \frac {5s^2 \plus{} r^2 \plus{} 4Rr}{32RS}\le \frac {3(R \plus{} r)}{4S} \Longleftrightarrow 5s^2 \plus{} r^2 \plus{} 4Rr\le 24R^2 \plus{} 24Rr\ (*)$ .

But in every triangle is true the following ineq. : $ s^2\le 4R^2 \plus{} 4Rr \plus{} 3r^2\implies 5s^2 \plus{} r^2 \plus{} 4Rr\le 20R^2 \plus{} 24Rr \plus{} 16r^2$ and from Euler's ineq. $ (*)$ is true.

[b]Second solution:[/b]

Obvious we have $ LHS\le \frac 12\left(\frac 1a\plus{}\frac 1b\plus{}\frac 1c\right)\equal{}\frac{ab\plus{}bc\plus{}ca}{2abc}$

Let's prove that $ \frac{ab\plus{}bc\plus{}ca}{2abc}\le \frac{3(R\plus{}r)}{4S}$ . This is equivalent with:

$ \frac{s^2\plus{}r^2\plus{}4Rr}{8RS}\le \frac{3(R\plus{}r)}{4S}\Longleftrightarrow s^2\plus{}r^2\plus{}4Rr\le 6R^2\plus{}6Rr\Longleftrightarrow$ $ s^2\plus{}r^2\le 6R^2\plus{}2Rr$

Using that $ s^2\le 4R^2\plus{}4Rr\plus{}3r^2\implies s^2\plus{}r^2\le 4R^2\plus{}4Rr\plus{}4r^2$ .

So we must prove that $ 4R^2\plus{}4Rr\plus{}4r^2\le 6R^2\plus{}2Rr\Longleftrightarrow 0\le R^2\minus{}Rr\minus{}2r^2\Longleftrightarrow 0\le (R\plus{}r)(R\minus{}2r)$ which is true .


There is a game involving $ 5$ boxes numbered $ 1, 2, 3, 4,$ and $ 5$ containing $ 0, 20, 30, 40,$ and $ 50$ coins respectively. The game has two players ($ A$ and $ B$) who take turns alternately. It begins with $ A$ taking some coins from some box $ n$ and moving them to box $ n\minus{}1$. (For example A could take $ 23$ coins from box $ 4$ and move them to box $ 3$ so that the boxes have $ 0, 20, 53, 17, 50$ coins respectively. Then $ B$ might take $ 2$ coins from box $ 2$ and move them to box $ 1$ so that the boxes have $ 2, 18, 53, 17, 50$ coins respectively.) You cannot move any coins that are in box $ 1.$ When all the coins have been moved to box $ 1$ the game is over. The player who makes the last move wins. Which player has a winning strategy and what is it?

[hide]
Let X denote a box with coins and O a box without. Box 1 doesn't matter because you can't move stuff from it, so denote it _.

A losing position means that you lose no matter what move you make. A winning position means that after one of your moves, your opponent is in a losing position. Define _ O O O O to be a losing position (because it means you just lost!)

Lemma 1.
_ X O O O is a winning position.

This is obvious - move everything from box 2 to box 1.

Lemma 2.
_ O X O X is a losing position. (as is _ O X O O)

Whichever player plays in this position (wlog player A) must move something from an odd-numbered box to an even-numbered box. Whatever he moves, player B moves all the coins from that even-colored box to the neighboring odd-numbered box. Then player B must make the last move, as A can only move to even-numbered boxes.

Now, I will show that player A wins. Player A should first move 20 coins out of box 4 into box 3. Next, as the coin layout will now be

0 20 50 20 50,

From now on, player A should keep the number of coins in boxes 2 and 4 equal. That is, if player B moves k coins from box 4 to box 3, then player A should move k coins from box 2 to box 1, and vice versa; and if player B moves k coins from box 5 to box 4, or box 3 to box 2, player A should move those same k coins to box 3 or box 1, respectively. This means that no matter what, B cannot make the number of coins in boxes 2 and 4 equal, because on his turn, they are already equal, and he must move coins, which will change the number of coins in either box 2 or 4 but not both. (I hope >.< otherwise it's a draw) As player B must eventually take the last coin(s) from one of boxes 2 or 4, player A will take the last coin(s) from the other, and will cause a position detailed in Lemma 2. Then he wins.

[/hide]


analytic geometry
probability
LaTeX
modular arithmetic
number theory
relatively prime

Find the smallest positive or zero integer $ z$ that makes the quantity
$ (3 \cdot (z + 3) ^{16} ) / 25$ a positive integer.

[hide]$ 2$[/hide]

$ S$ is the set of all points with coordinates $ (m,n)$ such that $ m$ and $ n$ are positive integers with $ m < 5$ and $ n < 5$ . Two points from $ S$ are chosen at random. What is the probability that the midpoint of the segment of the two points is also in the $ S$ ? Give your answer as a common fraction in lowest term.

[hide]$ \frac {1}{5}$[/hide]

Positive integers $ x$, $ y$ and $ n$ are each less than $ 20$ . If $ x^{2} - y^{3} = n$, find the median of all possible values for the sum $ x + y + n$ .

[hide]$ 20$[/hide]

[size=75]EDIT: $ \text{\LaTeX}$ $ \Rightarrow$ PROBLEM SLIGHTLY CHANGED.[/size]

EDIT Read the problem statement very wrong.  :huh:

[hide="1"]
\begin{eqnarray*}
\frac{3(z+3)16}{25}&=&\frac{48z+9}{25}\\
z&=&42
\end{eqnarray*}

This works because $ z\equiv2\pmod{10}$ for 48z+9 to be end with 5. It is impossible for 48z+9 to end with a zero. The smallest value is 42.
[/hide]

hunter34 the question say $ (z\plus{}3)^{16}$ not $ (z\plus{}3)16$,

anyway the way i did it was

[hide="Click to reveal non-hidden content for #1"]we know we need a factor of $ 25$ so we need at least 2 factors of 5 so $ 3\times{5^16}$ works so that would make $ z\equal{}2$. Just guess and check, if anyone has a less adhock way of doing please do tell[/hide]

[quote="AstroPhys"]The present statement is not the same as the original. hunter34 solved the original correctly.  :wink:[/quote]

Yes, I know.  Sorry, but I made a typo in the question, and then fixed it.  Anyway, can someone provide a solution for it now, please?

Thanks.

[hide="Number 1 when read correctly  :lol: "]
Since $ 3$ is relatively prime to $ 25$, $ \left( z\plus{}3 \right)^{16}$ must be a multiple of $ 25$.  The smallest value is $ z\equal{}2$.
[/hide]


Write $ \sqrt{25\minus{}12\sqrt{2}}$ in the form $ a\plus{}b\sqrt{c}$. Show your work.

[hide]$ \sqrt{25\minus{}12\sqrt{2}} \equal{} a\plus{}b\sqrt{c}$

$ 25\minus{}12\sqrt{2} \equal{} a^2 \plus{} 2ab\sqrt{c} \plus{} b^2c$

We can match up the terms, and we see that $ c\equal{}2$.

$ 25\minus{}12\sqrt{2} \equal{} a^2 \plus{} 2ab\sqrt{2} \plus{} 2b^2$

So we see that

$ 2ab \equal{} \minus{}12$

$ ab\equal{}\minus{}6$

and

$ a^2\plus{}2b^2 \equal{} 25$

Here if you try to substitute for $ a$ or $ b$, you end up with really ugly numbers so I'll leave it here for you.[/hide]

For most questions (take 2006 MC nats sprint #30 for example) you would be able to skip to mod. marc's last step (the $ a^2 \plus{} xb^2 \equal{} d$ step) and guess and check easily., but this problems a bit tough since you dont get integers.

[hide="This is just in case you dont know how to simplify radicals in radicals at all"]This was the 2006 mathcounts national number 30 sprint. number....30....sprint.....nationals, and I could solve it :wow:

The problem was essentially: simplify $ \sqrt {17 \plus{} 12\sqrt {2}}$into $ a \plus{} b\sqrt {c}$

Since 2 is prime, this problem is simpler. We know that the 17 term has to come from the $ a^2 \plus{} (b \sqrt {c})^2$ term. You can check this out be doing $ (1 \plus{} 2 \sqrt3)^2$ and writing out the complete factorization, noticing that the integer in the final answer results from the $ 1^2 \plus{} (2 \sqrt3)^2$

Since the 2 in the inside square root is prime, we can say that the sqrt(c) term is sqrt(2) (if the square root term wasnt prime, then we couldnt get the sqrt(c) term like that, and we would need a different approach. But since it is, we take the easy way). So :

$ a^2 \plus{} (b \sqrt {2})^2 \equal{} a^2 \plus{} 2b^2 \equal{} 17$

After a grand total of 5 seconds of guessing and checking (and knowing that since its MC, the a and b will be integers), we get b=2, a=3. Checking, $ (3 \plus{} 2 \sqrt2)^2$ it works.

How this made a number 30 escapes me.[/hide]


I was wondering whats an example of  closed sets $ U$ and $ V$ in $ \mathbb{R}^2$ such that $ U$ and $ V$ have no common point, but $ d(U,V)\equal{}0$?

d(U,V) is defined as inf of |x-y| where x belongs to U and y belongs to V.

the graphs of $ f(x) \geq \exp(x)$ and $ g(x) \leq \minus{}\exp( x)$.  Call these $ U,V$ respectively, and note that $ d(U,V) \leq 2 \exp(x)$, which goes to $ 0$ as $ x\to\minus{}\infty$, and hence $ d(U,V) \equal{} 0$, but $ U\cap V \equal{} \varnothing$.

\[ \begin{gathered}
  U \equal{} \left\{ {1,2,3,...} \right\} \hfill \\
  V \equal{} \left\{ {1 \plus{} \frac{1}
{1},2 \plus{} \frac{1}
{2},3 \plus{} \frac{1}
{3},...} \right\}. \hfill \\
\end{gathered}
\]

\[ \begin{gathered}
  U \equal{} \operatorname{epi}\frac{1}
{x} \hfill \\
  V \equal{} \left\{ {y \equal{} 0} \right\}. \hfill \\
\end{gathered}
\]

this is like exactly the same as my example, you know.


square RSTU is inscribed in the circle . what is the degree measure of arc ST

[hide=" click to reveal hidden content"]

$ 90^{\circ}$?

[/hide]

[hide]
The center of the circle is also the center of the square (the intersection of the diagonals, let's say), since the center of the square is also equidistant from R, S, T, U, and three points determine a circle, much less four. Parallelograms' diagonals bisect each other, and rhombuses' diagonals are perpendicular to each other, so $ \angle SOT \equal{} 90^\circ$.
[/hide]


I gave to the following a very nice solution, but impossible to find. So, I will let you find another solution:
    Prove that if a,b,c,d>0 verify abc+bcd+cda+dab=a+b+c+d, then
  sum sqrt((a^2+1)/2)<=sum a.

I will post my solution for this one since I'm absolutely sure that noone could find it.
  We will prove that (a^2+1)/(a+b)+(b^2+1)/(b+c)+(c^2+1)/(c+d)+(d^2+1)/(d+a)=a+b+c+d and after that Cauchy will do the rest.
  To prove the identity, just observe that (a+b)(a+c)(a+d)=a^2(a+b+c+d)+abc+bcd+cda+dab=(a^2+1)(a+b+c+d). Use similar identities and add them up.

It's unbelievable, but it's fact. Thank you Harazi
The indentity can be proven more naturally:
Sum (a^+1)/(a+b) - Sum a = Sum (1-ab)/(a+b)
But (1-ab)/(a+b) + (1-cd)/(c+d) = (c+d-abc-abd+a+b-cda-cdb)/(a+b)(c+d) = 0 and the second pair also add up to 0.

But the main thing is to get the identity, not to prove!

Namdung

The source is not at all unknown: I have discovered this inequality and all the others whose source wasn't indicated by me.


[hide]The LHS is always non-negative, so if  is a triple that produces a negative RHS then the inequality is trivially true.

So suppose that the RHS is always non-negative, that is  and the other cyclic permutations. Then a,b,c are the sides of a triangle, and our inequality is equivalent to

However, , so we have

where the inequalities follow from AMGM, definition of , and the definition of , respectively, and we're done.

[/hide]

This problem is thoroughly discussed here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122069[/url]

[quote=Ilthigore]Show that for all positive reals a,b,c

$(a^{2}+b^{2})^{2}\geq (a+b+c)(a+b-c)(b+c-a)(c+a-b)$[/quote]

If the RHS is negative or zero, then we are done. Otherwise $a,b,c$ form the sides of a triangle, and we want to show that

$$(a^2+b^2)^2 \ge 16\Delta^2$$
or
$$\frac{a^2+b^2}{4} \ge \Delta$$
which is true, because
$$\frac{a^2+b^2}{4} \ge \frac{ab}{2} \ge \frac{ab\sin C}{2} = \Delta$$


I wanted to attach it to my previous post. Sorry!

This inequality holds for any real $ n>0$.
Without loss of generality, we may assume that $ a_1\plus{}...\plus{}a_n\equal{}b_1\plus{}...\plus{}b_n$, when the inequality becomes
$ \sum (\frac {a_1}{b_1})^n(a_1\minus{}b_1)\ge 0$.
Since
$ (\frac {a_1}{b_1})^n(a_1\minus{}b_1)\ge a_1\minus{}b_1$,
the conclusion follows.

@Martin Bancos: Nice lemma! Seems like this could be extremely useful


Solve in real numbers the equation:

$ \frac{1}{x}\plus{}\frac{1}{x\plus{}2}\minus{}\frac{1}{x\plus{}4}\minus{}\frac{1}{x\plus{}6}\minus{}\frac{1}{x\plus{}8}\minus{}\frac{1}{x\plus{}10}\plus{}\frac{1}{x\plus{}12}\plus{}\frac{1}{x\plus{}14}\equal{}0.$

[quote="moldovan"]Solve in real numbers the equation:

$ \frac {1}{x} \plus{} \frac {1}{x \plus{} 2} \minus{} \frac {1}{x \plus{} 4} \minus{} \frac {1}{x \plus{} 6} \minus{} \frac {1}{x \plus{} 8} \minus{} \frac {1}{x \plus{} 10} \plus{} \frac {1}{x \plus{} 12} \plus{} \frac {1}{x \plus{} 14} \equal{} 0.$[/quote]

$ (\frac {1}{x} \plus{}\frac {1}{x \plus{} 14})\plus{}(\frac {1}{x \plus{} 2}\plus{}\frac {1}{x \plus{} 12})\minus{}(\frac {1}{x \plus{} 4}\plus{}\frac {1}{x \plus{} 10})\minus{}(\frac {1}{x \plus{} 6} \plus{} \frac {1}{x \plus{} 8})\equal{}0$
$ \Leftrightarrow (2x\plus{}14)(\frac{1}{x(x\plus{}14)}\plus{}\frac{1}{(x\plus{}2)(x\plus{}12)}\minus{}\frac{1}{(x\plus{}4)(x\plus{}10)}\minus{}\frac{1}{(x\plus{}6)(x\plus{}8)}\equal{}0$

Clearly $ x\equal{}\minus{}7$ is a solution. To find out the other solutions, we assume that $ x \neq \minus{}7$, then
 let $ y\equal{}(x\plus{}7)^2$ and the equation becomes:

$ (\frac{1}{y\minus{}7^2}\minus{}\frac{1}{y\minus{}1^2})\plus{}(\frac{1}{y\minus{}5^2}\minus{}\frac{1}{y\minus{}3^2})\equal{}0$
$ \Leftrightarrow \frac{3}{(y\minus{}7^2)(y\minus{}1^2)}\plus{}\frac{1}{(y\minus{}5^2)(y\minus{}3^2)}\equal{}0$
$ \Leftrightarrow y^2\minus{}38y\plus{}181\equal{}0$ with $ y \neq 1^2,3^2,5^2,7^2$

$ \Leftrightarrow y\equal{}19 \pm 6 \sqrt{5}$
$ \Leftrightarrow x\equal{}\minus{}7 \pm \sqrt{19 \plus{} 6 \sqrt{5}}$ or $ x\equal{}\minus{}7 \pm \sqrt{19 \minus{} 6 \sqrt{5}}$ as the only solutions other than $ x\equal{}\minus{}7$.


geometry
perpendicular bisector
geometry unsolved

Let $ ABC$ be a triangle with $ \angle BAC \equal{} 135^{\circ}$, altitudes $ AM$ and $ BN$. Line $ MN$ intersects the perpendicular bisector of $ AC$ at $ P$, let $ D$ and $ E$ be the midpoints of $ NP$ and $ BC$ respectively. Prove that $ ADE$ is a right isosceles triangle.

Denote I,F be the midpoints of AC,NC.
$ 2\angle MPI\equal{}90^o\plus{}2\angle ACB, \angle MIP\equal{}90^o\minus{}2\angle ACB\Rightarrow 2\angle MPI\plus{}\angle MIP\equal{}180^o\Rightarrow IP\equal{}IM$
Thus $ \angle APC\equal{}90^o\Rightarrow \angle ACP\equal{}45^o$
DF || BC, EF || IP, $ \angle DFE\equal{}\angle ICP\equal{}45^o, \angle DME\equal{}135^o$, so the quadrilateral DMEF is cyclic, hence five points A,D,M,E,F all lie on the circle whose diameter is AE,... And done.