Here's an interesting problem I created.  Ant #1 is climbing up and $60^\circ$ incline.  Ant #2 starts $10$ ft. from the vertex and walks at the same speed as ant #1.  How high will ant #1 be when it is directly above ant #2?

Just break up Ant #1's speed into $v_x$ and $v_y$. Then find the initial horizontal distance that separates them divide it by $\Rightarrow$ $v_x+\sqrt{v_x^2+v_y^2}$. Then use right triangles to find your answer.

[hide]
Triangle formed by vertex, ant #1, and ant #2 is 30-60-90.  Let $x=$ long leg.
$x=\sqrt{3}(10-x) \rightarrow x=15-5\sqrt{3}$
Thus height = $10\sqrt{3}-10$
This was done, fast.  I might have made an error.
[/hide]

is the circle there to show that they are both 10ft from the vertex?  if it is then, this is my answer.
[hide]
$\frac{20\sqrt{3}}{3}$
[/hide]

[hide="something"]When Ant #2 goes 20/3 feet, it will be directly under Ant #1. Soo... Ant #1 will be 10/sqrt(3) feet in the air.[/hide]

[hide="My answer"]

I figured that the horizontal speed that Ant 1 covers is half of that of Ant 2 because cos 60 = .5.

That means Ant 1 moves half a foot for every foot that Ant 2 moves.

Let's say that each ant's speed is 1 ft/s. The horizontal distances will be equal when time = $\frac{20}{3}$ seconds.

In that time, the ant will have moved a vertical distance of $\frac{20}{3} \cdot sin 60$ which is equal to $\frac{10\sqrt{3}}{3}$ feet.

Is that right?
[/hide]

[quote="d343seven"]is the circle there to show that they are both 10ft from the vertex?  if it is then, this is my answer.
[/quote]

Yeah, I guess the circle is useless.  The original problem would have only given the radius of the circle as shown below, and the answer would have been twice the radius, or 20.

Here's my contrived solution:
Since one ant travels towards the vertex and the other ant travels away from it at the same speed, the sum of their distances from the vertex should always be 20 ft.
$h\cot\theta+h\csc\theta=20$
$h(\cot60^\circ+\csc60^\circ)=20$
Using the identy $\cot\left(\frac{1}{2}\theta\right)=\cot\theta+\csc\theta$,
$h\cot30^\circ=20$
$h=\frac{20\sqrt{3}}{3}$

is the circle useless?  i used the circle to see that the 2 ants are both 10 away from the vertex, since they are at tangent points.
so then i said the hypotenuse was 10+x and the side facing the 30 deg angle was 10-x, since they travel the same distance in the same time.  then i solved for x and got the height from there.

[quote="juggle7"]
Yeah, I guess the circle is useless.  The original problem would have only given the radius of the circle as shown below, and the answer would have been twice the radius, or 20.
[/quote]


[quote="t0rajir0u"]For $x, y, z > 0$, show that

(a)  $\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} \ge \frac{y}{x} + \frac{z}{y} + \frac{x}{z}$

(b)  $\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} \ge \frac{x}{y} + \frac{y}{z} + \frac{z}{x}$[/quote]

[hide="a"]let $a=\frac{x}{y}$, etc, then we have

$a^2+b^2+c^2\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ for $abc=1$, multiply the RHS by $1=abc$ to get
$a^2+b^2+c^2\ge ab+bc+ca$...trivial[/hide]

[quote="4everwise"]These two problems are like exactly the same...  :huh:

[hide="a)"]Apply AM-GM in 3 seperate cases.

$\frac{x^2}{y^2}+\frac{y^2}{z^2}\ge 2\frac xz$, and similarly for the others. Adding and dividing by 2 gives the result. [/hide][/quote]

I actually had a harder time with the second, but I did find a one-liner for both.  Nice solution.  :)

For part (b): Let $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$, then the inequality becomes:
\[ a^2+b^2+c^2 \geq a+b+c  \]
$a+b+c = (a+b+c) \cdot \sqrt[3]{abc} \leq \frac{(a+b+c)^2}{3} \leq a^2+b^2+c^2$
Equality holds if and only if $a=b=c$.

[quote="t0rajir0u"]Yes, that's more or less what I came up with.  :)  (The book's solution involves clearing the denominators, so I wasn't very satisfied with it.)[/quote]

which book?

[i]Problem-Solving Strategies[/i], by Arthur Engel.  Very good Olympiad preparation, but I thought this was an easier problem so I put it here.

[quote="shadysaysurspammed"][quote="maokid7"]for part b couldnt u use Cauchy's?[/quote]

No you can't...(unless I'm doing something wrong :? )

$(\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2})(1^2+1^2+1^2)\geq ( \frac{x}{y} + \frac{y}{z} + \frac{z}{x} )^2$[/quote]

You're one step away.  ;)


analytic geometry
graphing lines
slope
LaTeX

[quote="Cyberman"]5(2-x)/3 - 1 = x + 5

Answer is -1

I've tried out this problem many times and I just don't have clue as to how I'm suppose to go out solving for x. I understand slope intercept form so i tried using the reciprocal in there somewhere but I didn't get -1. I usually obtained something like 2 or some other weird number.

Could someone show me the steps?[/quote]
[hide]
5(2-x)/3 - 1 = x + 5

Multiply expression in parentheses by 5/3:

10/3 - 5x/3 -1 = x + 5

Add constant terms on the left hand  side of the equals:

7/3 - 5x/3 = x + 5

Subtract 7/3 from both sides:

-5x/3 = x + 8/3

Subtract 3x/3 from both sides:

-8x/3 = 8/3

Multiply both sides by -3/8:

x = -1.
[/hide]

[quote="anirudh"]$\text{Did you mean}$

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

$\frac{10-5x}{2}=x+5$

$10-5x=2x+10$

$\therefore x=0$

$or$

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

$\frac{5(2-x)}{3}=x+6$

$5(2-x)=3x+18$

$10-5x=3x+18$

$8x=-8$

$x=-1$

$\text{P.S. You should have written it like this. Well I am not using latex}$

[5(2-x)/3] - 1=x+5 if u meant number 2[/quote]
i think he did mean that.  He just didn't use $\LaTeX$ ;)

[hide][quote="Cyberman"]5(2-x)/3 - 1 = x + 5

Answer is -1

I've tried out this problem many times and I just don't have clue as to how I'm suppose to go out solving for x. I understand slope intercept form so i tried using the reciprocal in there somewhere but I didn't get -1. I usually obtained something like 2 or some other weird number.

Could someone show me the steps?[/quote]

$
\displaystyle \frac{5(2-x)}{3} - 1 = x+5
\displaystyle 5(2-x) - 3 = 3x+15
\displaystyle 10-5x-3 = 3x+15
\displaystyle -8x = 8
\displaystyle x = -1
$

$ \text{\LaTeX} $ is pretty easy (at least the basics are).  then you just fiddle around and scan through the Guides for something you don't know, and you become pretty good at it.  I'm just starting.  I don't really read the whole guide and try to memorize it or anything.  It just is a lot easier to read
[/hide]

[quote="Cyberman"]5(2-x)/3 - 1 = x + 5

Answer is -1

I've tried out this problem many times and I just don't have clue as to how I'm suppose to go out solving for x. I understand slope intercept form so i tried using the reciprocal in there somewhere but I didn't get -1. I usually obtained something like 2 or some other weird number.

Could someone show me the steps?[/quote]

[hide]$\frac{5(2-x)}{3}-1=x+5$

$5(2-x)-3=3x+15$

$10-5x-3=3x+15$

$7=8x+15$

$8x=-8$

$x=-1$[/hide]

For that problem, all you needed to know was 1 $\LaTeX$ command, \frac. For most equations, you just enclose your question (or answer) in $ signs or double $ signs for centering.

For example:

\[5^2+3^2\]

or

$5^2$

For fraction:

[code]\frac{numerator}{denominator}[/code]

With the dollar signs


geometry
trigonometry
geometric transformation
reflection
geometry unsolved

In $\triangle ABC$, $\angle ABC=90^\circ$, and $D$, $G$ are two points on $CA$. Let $E$, $F$ be the projection of $A$, $G$ on $BD$ respectively. Suppose $BE=EF$. Show that $\angle ABG=\angle DFC$.

We will prove that if $\angle ABG=\angle DFC$ and $GF\perp BF$, then the median $AM$ of $\triangle ABG$ is orthogonal to BF.

Let H be the orthocenter of $\triangle BCG$, then $\triangle BHC\sim\triangle AMB$, thus it suffices to show that BD coincides with the mediam BM' of $\triangle BHC$. Let X be the point of intersection between BD and the perpendicular of C to AC. Simple angle chase would yield $\angle GXC=B$, thus $CX=b/\tan B=BH$, done.

One member from Hong Kong came up with an elegant solution just aftre the end of the competition :( , which in my opinion is even more beautiful than the official one.
Reflect the point $F$ with respect to $AC$ to $F'$. Let $\angle ABG =\alpha$ and $\angle FBC=\beta$. It is obvious that $\beta =\angle BAE=\angle FAE=\angle AFG=\angle AF'G$. Hence $\angle ABD+\angle AF'D=(90^\circ-\beta)+(90^\circ+\beta)$, implying $A,B,D,F'$ are concyclic, and so $\angle F'AD=\angle F'BD$. Now $\angle F'GC=\angle F'AD+\beta=\angle F'BD+\beta=\angle F'BC\Rightarrow F',G,B,C$ are concyclic. Therefore, $\angle DFC=\angle DF'C=180^\circ-\angle DF'G-\angle GBC=180^\circ-90^\circ-(90^\circ-\alpha)=\alpha=\angle ABF$.


quadratics
algebra
polynomial
Rational Root Theorem

I came across this different way of factorising some quadratics that I have never heard of before, and thought it was pretty clever.

When you are trying to factorise $x^2 + bx + c$, you look for two numbers that multiply to give c and add to give b. But what happens if we have $ax^2 + bx + c$? The usual way (that I hear anyway) would be to find numbers that multiply to a and numbers that multiply to c and then testing all the possibilities to get the b term right. After a while you get fast at them - but how about this for a nicer way.

Look for two numbers that multiply to ac and add to b. Thats no harder than factorising a simple quadratic. Lets say they are p and q.

Then just write it as $(ax^2 + px) + (qx + c)$ and factorise each of those brackets separately. You will end up having a common linear factor, and the whole thing will factorise.

Example:

Factorise $12x^2 + x - 6$.

We want two numbers that multiply to -72 and add to 1. They are 9 and -8. So:
$12x^2 + x - 6 \\
= 12x^2 + 9x - 8x - 6 \\
= 3x(4x + 3) - 2(4x+3) \\
= (3x-2)(4x+3)$.

Or, if you did it the other way around:
$12x^2 + x - 6 \\
= 12x^2 - 8x + 9x - 6 \\
= 4x(3x - 2) + 3(3x-2) \\
= (4x+3)(3x-2)$.

It may sound a bit hard, but its no harder than factorising a simple $x^2 + \ldots$ quadratic. And it always works. Quite interesting.

Here is a rather "magical" method for factoring quadratics.

Same example: $12x^2 + x - 6$

(1) Write down the parentheses and "split" the $x^2$ term.

Use the leading coefficient [b]twice:[/b] $(12x\qquad)(12x\qquad) $

(2) Obtain $ac$ and factor the product.
Since the last term is negative, factor $ab = 72$ into two parts
whose [u]difference[/u] is 1, the 'middle' coefficient. [b]**[/b]
We find that $72 = 8\times9$ is the desired pair.

Since we want $+1$, we will use $-8$ and $+9.$
Insert them into the parentheses: $(12x - 8)(12x + 9)$

(3) Factor out all common factors: $4(3x - 2)\;3(4x + 3)$

. . . and discard them! . . .  $(3x - 2)(4x + 3)$ . . . [i]There![/i]

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

[b]**[/b] If the last term is positive, factor $ac$ into two parts
whose [u]sum[/u] is the middle coefficient.

TripleM's way was the way my Algebra II book taught factoring higher numbers.  They called it "splitting the middle term".

Actually, both methods are the same, but Sorobans skips out the messy factorisation step and makes it a lot nicer and much more 'magical' :) (I actually came across the same final formula of $(ax+p)(ax+q)/a$ when trying to prove the first one).

Here's another way to look at it which I think is less magical.  Apologies if it is the same as what was said above.

We want to factor $ax^2 + bx + c$.  We multiply the quadratic by the constant $a$.  That gives $(ax)^2 + b(ax) + ac$.  We can view that expression as a quadratic in $ax$.  Thus we have reduced the problem to the case when the leading coefficient is 1.

This reminds me of an observation I once made (probably well known) related to the Rational Root Theorem.  For example, suppose that $a$, $b$, and $c$ are integers.  The Rational Root Theorem says that a rational root of the quadratic (or higher polynomial) must be the quotient of a factor of $c$ divided a factor of $a$.  The only problem is that you may get the same candidate root many times.  For example, if $a = 2$ and $c = 2$, then you end up with both 1/1 and 2/2.  Instead, we can phrase the theorem as saying that rational roots must be a factor of $ac$ divided by $a$.  That way there are no duplicates.


The combined weight of three basset hounds is $ 185$ pounds. The two smaller dogs weigh the same. The difference between the larger weight and the smaller weight is $ 20$ pounds. How many pounds does the largest dog weigh?

Let each of the small dogs weigh x pounds. Then the large one weighs x+20 pounds, and all three weigh 3x+20 together. This must be 185, so 3x+20=185, so 3x=165 and x=55. But we are looking for x+20, so the answer is 55+20=75.

Since you would like to know how much the larger dog weighs, it would be faster to make x the larger dog so that 3x-40=165, 3x=225, and x=75. This way, you don't need to go through the extra step of adding 20 to the first answer. :agent:

Let the three basset hounds weigh $a$, $a$, and $b$ pounds, where $a < b$. We have the two equations \begin{align*}
2a+b&=185\\
b-a&=20
\end{align*} From the second equation, we have $a=b-20$. Substituting this in to the first equation to eliminate $a$, we have $2(b-20)+b=185 \Rightarrow b=75$. Thus, the largest dog weighs $\boxed{75}$ pounds.

(This is the Alcumus answer, not mine, i'm not stealing)


algebra
polynomial
irrational number
Rational Root Theorem
algebra unsolved

$P$ is a real polynomial such that if $\alpha$ is irrational then $P(\alpha)$ is irrational. Prove that $\deg P \leq 1$.

Let P(x) irrational, when x irrational. If deg(P(x))=n>0, then exist (n+1) big rational numbers $r_i: P(x_i)=r_i$. Therefore $x_i$ are rational $\Longrightarrow P(x) \in Q[x]$.  If P(x)-a (a - rational) had irrational roots we have contradition. For all rational polinoms with deg(P(x))>1 exist a: $P(x)-a$ had irrational roots. Therefore P(x)= irrational const or ax+b, a and b rational.

[quote]For All $P(x) \in Q[x]$ there exist a rational number like $R$ such that ,
$P(x)-R$ has an irrational root [/quote]
Obviously this is Equal to the problem .
Why there exist such a rational number ?

Let $F=\{x: \exists a\in \mathbb{R}\setminus \mathbb{Q} s.t. P(x)-a=0\}$. We assume that $F\subset \mathbb{Q}$.

Define a map which maps every $a\in \mathbb{R}\setminus \mathbb{Q}$ in the set $\{x_1,x_2,...x_n\}$ which are the roots of $P-a$. We can see that this map is injective. Hence there is an injetion from $\mathbb{R}\setminus \mathbb{Q}$ into $\mathbb{Q}^n$, but the first is not numerable, and the second is. Contradiction.

[b]xirti[/b], the problem stipulated that the value of $P$ at any irrational number is irrational, not rational.

We first claim that $P$ must have rational coefficients. Without loss of generality, we may let the leading coefficient of $P$ be positive. It follows that for all sufficiently large integers $n$, we can find some real $q$ for which $P(q) = n$. $q$ must be rational, or else $P(q) = n$ would be irrational. Hence, by the Lagrange interpolation formula, $P$ must have rational coefficients.

Let $Q(x) = n P(x)$, where $n$ is a positive integer chosen so that $Q$ has integer coefficients. Let the leading coefficient of $Q$ be $c$, and let its constant term be $d$. By assumption, for all sufficiently large primes $p$, we can find some positive rational number $r$ such that $Q(r) = p + d$. By the rational root theorem, all rational roots of the polynomial $Q(x) - (p-d)$ must be of the form $\frac{\pm p}{s}$, where $s$ is a divisor of $c$. Hence, $r \geq \frac{p}{c}$. On the other hand, if $\deg Q > 1$, then $0 = Q(r) - (p-d) \geq Q(\frac{p}{c}) - (p-d) \geq O(p^2)$, which is clearly impossible. Hence, the degree of $P$ must be less than or equal to 1.


factorial calculus geometry function trigonometry logarithms search

well, how many ways can you arrange nothing?  :D

this has been proven before too on AoPS... look[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=factorial+0&t=38005[/url]

it's just a convention because you can answer 'literally' the above question by dangvh depending on how you 'philosophically' think: you can say it's zero ways because there is nothing that you can arrange and you can say it's one because arranging nothing is still one way.

the convention most widely accepted is $0!=1$

even more interesting, what's 1.5!    :)

[quote="Barnacle"]even more interesting, what's 1.5!    :)[/quote]

it says 1.3293... on my calculator

if there are 2 seats and 2 people, there are 2! ways to arrange them

if there is 1 seat and 1 person, there is 1! ways to arrange it

so if there are 0 people and 0 seats there are0! wasys to arrange wich is 1 so 0!=1 :P

[quote="Barnacle"]even more interesting, what's 1.5!    :)[/quote] is there a way to perform that operation?

$\left( \begin{array}{c} n\\ 0 \end{array} \right) = \frac{n!}{n!0!}=\frac{1}{0!}$

Just think about that.  ;)

Any way for the original question:
To get  (n-1)!  all you have to do is to divide  n! by n.  Now we all know  1! is 1 so to get
0! divid  1! by 1 and you get 1.

Hope this sounds a little sensible..

[quote="Gyan"][quote="jli"][quote="Barnacle"]even more interesting, what's 1.5!    :)[/quote]

it says 1.3293... on my calculator

 [/quote]

Well if you really want to write it  nicely  is also  $ \frac {3 \sqrt \pi} {4}   $
Check it out and enjoy![/quote]
how'd you figure that???  :o

Dangvh -This is something which people learn   when they  take  a few years of calculus :)  - that
$ (1/2)! $ is $ \frac {\sqrt \pi } {2} $ and   (3/2)! is  nothing but $ (3/2) ((1/2)!) $

Basically  another way to "define" a factorial (or a generalized factorial) is define it as  total area  (from 0 to infinity) under a specific curve ( $ x^n  e^{-x} $ ) . One can show that if one defines a function like this,
$n! = n(n-1)!$ and  $1! = 1$ and thus for natural numbers the old definition and this definition is  the same.

Hope this helps.

Wow. I never knew factorials were so complicated. (After all, I've been using them since Kindergarten! lol)  :D

But how do you calculate stuff like say , $1.3!$ without using a calculator ?

[quote="236factorial"]Why does 0! equal 1? At least it does on my calculator.[/quote]

As others have already said, it follows from the idea that there is only one way to arrange nothing. To put it more intuitively though, "When you have no choice, there is only one choice to make."

Proof:

$\binom{n}{n}=\frac{n!}{n!(n-n)!}=1$

$\Rightarrow \frac{1}{0!} = 1$

$\Rightarrow 1 = 0!$

[quote="Barnacle"]even more interesting, what's 1.5!    :)[/quote]

The answer is: $\Gamma (2.5)$ , the proof of which is beyond the scope of this forum.

Read at on your own risk .. Comments which follow may be beyound the scope ..:)
You can safely omit   the rest ...
[hide][quote="Dr. No"]The answer is: $\Gamma (2.5)$ , the proof of which is beyond the scope of this forum.[/quote]

True  but $\Gamma $ function and  $!$ are related as $\Gamma (n+1) = n!$  so it  is of little help.  (Like saying I live  one house north of the house which is one house  south of my house  :)

Karthik: [quote]But how do you calculate stuff like say , 1.3! without using a calculator ?[/quote]

To get the right perspective, without a calculator how do you get stuff like say $\sin 1.3 $ or  $\log 1.3 $ .. Not that easy..Not for new kids now a days .. unlike us who had no calculators ...who had to walk 5 miles uphill both ways to scholol etc.. :)

Anyway once you know (or know how to calculate) factorials of numbers between  0 and 1 (say .3) you can calculate it for any other number (say 1.3)  etc..

One easy way is to use striling approximation to calculate  $10.3!$ and then calculate $1.3!$

Anyway apart from the case when n is integer, or half integer, there are no nice formulas...
[/hide]

[quote="Gyan"]Read at on your own risk .. Comments which follow may be beyound the scope ..:)
You can safely omit   the rest ...
[hide][quote="Dr. No"]The answer is: $\Gamma (2.5)$ , the proof of which is beyond the scope of this forum.[/quote]

True  but $\Gamma $ function and  $!$ are related as $\Gamma (n+1) = n!$  so it  is of little help.  (Like saying I live  one house north of the house which is one house  south of my house  :)

[/hide][/quote]

$\Gamma (n) =(n-1)!$ is a result, not a definition. On the other hand, $n!$ is only defined on the non-negative integers. It follows that $1.5!$ is nonsense. The answer to the question, "What is 1.5! ?" is either a perplexed shrug or $\Gamma (2.5)$.

[quote="Dr. No"]
..$n!$ is only defined on the non-negative integers.  It follows that $1.5!$ is nonsense. The answer to the question, "What is 1.5! ?" is either a perplexed shrug or $\Gamma (2.5)$.[/quote]

You have right to belive that.   I  guess some will even say $2^{1.3} $ or $\sqrt {-1.3}$ is "nonsense".  Depends on how one  defines   it and what makes  "sense" to one.

On the other hand  many calculators, math references, even just typing  "1.3!"  in google and hitting enter [url=http://www.google.com/search?hs=wpt&hl=en&lr=&safe=off&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&biw=1024&q=1.3%21&btnG=Search](eg here)[/url] will give the answer what  Karthik was wondering about.

We are going off the subject, so I will stop here, but for this audience it is important to know that  "generalization"  (  Eg   $a^n$ when n is integer is easy to understan but can $n$ be  a  fraction ? - questions like that) is fun and important part   of math. .. in many cases that is how, interesting concepts in  math evolved.

Hope that helps.

[quote="Gyan"]
On the other hand  many calculators, math references, even just typing  "1.3!"  in google and hitting enter will give the answer what  Karthik was wondering about.
Hope that helps.[/quote]

Well , I do know that if I say 1.3! on google , it gives me 1.16671191 ...Fortunately or unfortunately , I dont belong to the set of people who just accept whatever a calculator says...  :(
Well , how exaclty do you solve , say 1.8! when you have no electronic devices at all to help you ?  :o

Do you know calculus? It's improper integrals or something like that. http://mathforum.org/library/drmath/view/54540.html

[quote="chess64"]When do you get to learn this $\Gamma$ stuff?[/quote]

Advanced calculus in college ... that's why I said it was beyond the scope of this forum. But since I mentioned it, I'll make a couple of observations so no one gets confused.

The $\Gamma$ function is a calculus integral that has the result that when $n$ is an integer $\Gamma (n) = (n-1)!$. However $\Gamma (z)$ can be calculated for any $z$, real or complex, integer or not. (*Provided $\Re (z) > 0$) Consequently, when you enter 1.3! into your calculator, the calculator doesn't give you an error, as it should because factorial is only defined for integers; instead, it computes $\Gamma (2.3)$, because this is how the calculator obtains the factorial of an integer number anyway.

[quote="chess64"][quote="Dr. No"][quote="chess64"]:huuh: My TI-83+SE gives me an error...[/quote]

For what number?[/quote]

I type in $1.3!$, it says $\text {ERR:DOMAIN}$ (domain error)

EDIT: $1.5!$ works, though :)[/quote]

Weird ... though I believe I know why. Gamma for half arguments (0.5, 1.5, 2.5, etc.) can be made into a Gaussian integral for which the solutions are well known. In other words it's a look it up and spit it out sort of thing. For other arguments actually calculating the integral is very difficult. Rather than pop a transistor trying to figure it out it just gives you an error.

[quote="chess64"]:huuh: My TI-83+SE gives me an error...[/quote]

So will, I guess many ordinary calculators.  But  many (I know of) good calculators (Eg TI-86, 89) , maple, math references , google (BTW  math interface there was written by, I am told , a  former IMO)  should  be okay but that does not mean all calculators will  compute it fine.  For example,  maple or google would calculate $\sqrt {-1} $  while many other calculators will give an error.   (But then there are  calculators which  will give an error if you want to do something like $2^{1.3}$too!)

For most  of the math word,  it is perfectly acceptable to use '!' for non-integers (even  for complex numbers) .   $\Gamma$ is closely related. Historically $\Gamma$ was  popular but now a days  one uses '!'   too. Of course  in practice  it makes no difference because one is related to other.

For Karthiks question - how do  you calculate it (If your calculator does not do it for you) .. see my previous note.  for some hints ..
 One  method which many use is  the well known  approximation
$n! $ as appox equal to $ (\sqrt {(2n+1/3)  \pi})   n^n   e^{-n} ) $ for large n ( you will get a few significant figures  correctly  even when n is as small as 10)
(If n is small, just use the formla for  n+10 and then divide  by (n+1)(n+2)...(n+9) etc  )
 Again, for  more details you may like to look up strilings formula... or other standard math books...

Anyway   it is getting too  advance for this level .. if people are curious they may  ask this  question in  more advanced forum...

[quote="Gyan"]$n! $ as appox equal to $ (\sqrt {(2n+1/3)  \pi})   n^n   e^{-n} ) $ for large n ...
Anyway   it is getting too  advance for this level .. if people are curious they may  ask this  question in  more advanced forum...[/quote]
yes yes yes, please, please do ask in advanced forum. the only thing i understood is the above formula, everything else i'll leave that for now

Okay, I think I have to step up now.

For further discussion about this matter, please discuss in the upper forum.

If another post is made including advanced material after this, I'll have to lock this topic.

To Gyan and Dr.No, thank you very much for answering and spending time in getting started forum. $1.5!$ is a kind of question that I can't answer (since I'm basic to calculus).


$ ABCD$ is a quadrilateral such that $ \angle CBA \plus{} \angle CDA \equal{} 270$ deg, $ \angle BAD \equal{}\angle BCA\equal{} 30$ deg. $ m$ and $ n$ are the simetric lines  of the diagonal $ AC$ with respect the sidelines $ AD$ and $ AB$. $ P$ and $ Q$ are the orthogonal projections of $ B,D$ on $ m,n$ respectively. If $ M \equal{} BP \cap DQ$. Show that  $ MP \equal{} MQ$.

By direct calculation we get AC the bisector of the angle <BCD, AB and AD external bisectors of angles <CBP and <CDQ, that is, the quad BCDM has all its sides tangent to a circle centered A, hence P and Q are the contact points of BP and DQ with this circle, so MP = MQ.

Best regards,
sunken rock


Kiyoto picks a prime positive integer less than 10 (with all choices equally likely). Konstantin picks an odd positive integer less than 10 (with all choices equally likely). Compute the probability Kiyoto and Konstantin pick the same number.

[hide]1,2,3,5, and 7 fits Kiyoto's specification. 1,3,5,7, and 9 fits Konstantin's. So 1,3, and 5 can be picked by both. Since the total number of possibility is 25 (5*5), the answer is 3/25[/hide]

It seems too easy...hmm...was it a trick question?

Aaaaaah! That always gets me!!!   :evil:  :cry:  :-x

Anyways, so the answer is [hide]3/20[/hide]?


Given $ a,b,c \ge 0$and $ 0 \le k \le 1$.Prove that:
$ \frac{(a\plus{}b)^2}{c^2\plus{}kab}\plus{} \frac{(b\plus{}c)^2}{a^2\plus{}kbc}\plus{} \frac{(c\plus{}a)^2}{b^2\plus{}kca} \ge \frac{12}{k\plus{}1}$
For k=1,we have:
$ \frac{(a\plus{}b)^2}{c^2\plus{}ab}\plus{} \frac{(b\plus{}c)^2}{a^2\plus{}bc}\plus{} \frac{(c\plus{}a)^2}{b^2\plus{}ca} \ge 6$(Darij grinberg)

We only need to prove with $ k\equal{}1$. SOS solved it easily.
PS: [hide]Anh con` co' 1 loi` giai? khac' = 1 tieu chuan? SOS cu~ cua? anh. Tiec la` bay gio` anh van~ chua post tieu chuan? y' len dc :P[/hide]

[quote="secrets"]Can you explain,nguoivn,why do you prove for k=1?
[hide]Em khong hieu? y' anh choa lam',tai. sao lai la chung minh cho k=1 a.?[/hide][/quote]
Actually As $ k\le 1$ thus all other values of $ k$ except 1 are less than 1.
$ LHS\equal{}\sum\frac{(a\plus{}b)^2}{c^2\plus{}kab}\ge \frac{12}{1\plus{}1}$
Completes our proof; because in all places the $ k$ is in the denomiantor$ \implies$ RHS is always maximum if $ k\equal{}1$.
I hope I was able to explain it. :blush:  :)

[quote="Potla"][quote="secrets"]Can you explain,nguoivn,why do you prove for k=1?
[hide]Em khong hieu? y' anh choa lam',tai. sao lai la chung minh cho k=1 a.?[/hide][/quote]
Actually As $ k\le 1$ thus all other values of $ k$ except 1 are less than 1.
$ LHS \equal{} \sum\frac {(a \plus{} b)^2}{c^2 \plus{} kab}\ge \frac {12}{1 \plus{} 1}$
Completes our proof; because in all places the $ k$ is in the denomiantor$ \implies$ RHS is always maximum if $ k \equal{} 1$.
I hope I was able to explain it. :blush:  :)[/quote]
Dear friend,but $ 6 \ge \frac{12}{k\plus{}1}$ is not true
Try k=0,5.We get $ 6 \ge \frac{12}{1,5}\equal{}8$


function
topology
real analysis
real analysis unsolved

let f from (o,1) to R  be differentiable function with f'>0,let A in (0,1) s.t. closure(A) in (0,1) as well,show
 f(closure(A))=closure(f(A)). :)

By the inverse function theorem $f^{-1}$ is also differentiable, hence continuous. Since $f$ is a homeomorphism, all topological concepts like closure, interior, boundary.. are well-behaved under $f$.


I want to get better at minimum/maximum questions, and I was wondering if you guys could give me complete solutions to the following questions.

1. If $ x$ and $ y$ are two non-zero real numbers which satisfy the equation $ x^2 \plus{} y^2 \equal{} (x \plus{} y)^3$ then what is the minimum of $ x \plus{} y$?

2. Suppose that $ x$ and $ y$ are in $ (2, \minus{} 2)$ [b](what does that even mean)[/b] and $ xy \equal{} \minus{} 1$. What is the minimum value of $ \frac {4}{4 \minus{} x^2} \plus{} \frac{9}{9 \minus{} y^2}$?

Thanks.

[hide="#1"]We have by Cauchy-Shwartz inequality, that

$ (x^2 \plus{} y^2)2 \geq (x \plus{} y)^2$

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

Substitute $ x^2 \plus{} y^2 \equal{} (x \plus{} y)^3$.

$ (x \plus{} y)^3 \geq \frac {(x \plus{} y)^2}{2}$

$ x \plus{} y \geq \frac {1}{2}$

Therefore the minimum is $ \boxed{\frac {1}{2}}$.[/hide]

[hide="#2"]I haven't figured this one out yet but $ (\minus{}2, 2)$ means that $ \minus{} 2 < x , y < 2$. This is called interval notation.[/hide]

[quote="modularmarc101"][hide="#2"]I haven't figured this one out yet but $ (2, \minus{} 2)$ means that $ \minus{} 2 < x , y < 2$. This is called interval notation.[/hide][/quote]
No, it does not.

$ (\minus{}2,2)$ means $ \minus{}2<x, y<2$.

Are you sure you didn't typo, Thunder365?

Same thing (not really), but we get the point that its interval notation for that.

[hide="Solution"]This is equal to $ f(x) \equal{} \frac4{4 \minus{} x^2} \plus{} \frac {9x^2}{9x^2 \minus{} 1} \equal{} 1 \plus{} \frac4{4 \minus{} x^2} \plus{} \frac1{9x^2 \minus{} 1}$. Note that $ f(x) \equal{} f( \minus{} x)$ so we can WLOG restrict ourselves to $ 0\le x < 2$.

But for $ \minus{} 2 < y \equal{} \minus{} \frac1x < 2$, we must have $ x > \frac12$, so $ \frac12 < x < 2$.

$ f(x) \equal{} 1 \plus{} \frac4{4 \minus{} x^2} \plus{} \frac1{9x^2 \minus{} 1} \equal{} 1 \plus{} \frac {35}{37 \minus{} \left(9x^2 \plus{} \frac4{x^2}\right)}$.

After some stuff we find that stuff works when $ x \equal{} \sqrt\frac23$, which gives $ \min f(x) \equal{} \boxed{2.4}$.[/hide]

[quote="FantasyLover"][quote="modularmarc101"][hide="#2"]I haven't figured this one out yet but $ (2, \minus{} 2)$ means that $ \minus{} 2 < x , y < 2$. This is called interval notation.[/hide][/quote]
No, it does not.

$ ( \minus{} 2,2)$ means $ \minus{} 2 < x, y < 2$.

Are you sure you didn't typo, Thunder365?[/quote]

Oops, yeah, x,y are in -2,2, not 2,-2. sorry. Math154 is still right, though.

Thanks!


I memorized those sequence formulas before the startup lolz. I don't think it works for factorials.

Memorizing Formulas is Fun !  :wink:

I hardly knew any because hardly any previous experiences with MAth Team till this year in Geometry. and boy am i hooked up!

AOPS ROCKS!

[quote="7h3.D3m0n.117"]i no the trick for the problem 100+99+98+...+3+2+1.
and its 5050 for the answer

is it possible to apply the same thing to the question 100!+99!+98!+...3!+2!+1!

now i no it is rather quite large because i encountered the following problem

what are the last two digits of the sequence 100!+99!+98!+...3!+2!+1!?[/quote]
Well, you don't need the entire thing, just the last two digits.
You will notice that the last 2 digits of 10! are 00. So all of the last two digits after that will end in 00 as well. So for this all you need to do is find the sum of the last to digits of
$9!+8!+7!+6!+5!+4!+3!+2!+1!$

[quote="bpms"][quote="i_like_pie"][quote="7h3.D3m0n.117"]i no the trick for the problem 100+99+98+...+3+2+1.
and its 5050 for the answer

is it possible to apply the same thing to the question 100!+99!+98!+...3!+2!+1!

now i no it is rather quite large because i encountered the following problem

what are the last two digits of the sequence 100!+99!+98!+...3!+2!+1!?[/quote]
I did a problem similar to this in MC today. :)

As [b]bpms[/b] said, you need the sum of the first 9 factorials.

$9!+8!+7!+6!+5!+4!+3!+2!+1!=409113.$ The last two digits of the first 100 factorials are $1$ and $3.$[/quote]
A couple things. Why did you bold my name? Just wondering.
Also, you don't need to even find the complete sum, you just need the last two digits of each factorial, and from five up you know that the units digit is 0.[/quote]
I just felt like making it bold. Looks better(?).

You still have to multiply everything, so might as well write the whole thing.


vector
topology
Functional Analysis
linear algebra
real analysis
real analysis unsolved

Ahh, bombardment with technical banach space questions.

Suppose $ M$ is a weak* closed subspace of $ X^*$. Then $ M$ is the dual of a quotient of $ X$.

The idea seems to be to let ${ K = \vphantom{M}}^{\perp}{M}$. Then ${ (X/K)^* = (\vphantom{M}}^{\perp}{M})^\perp$

To show that this is equal to $ M$ we would like for every $ \ell \in X^*$ not in $ M$ to find an $ x$ that is annihilated by everything in $ M$ but not $ \ell$. Since $ M$ is closed I know how to find an $ x^{**}$ that does the job, but not an $ x$

$ M$ is not just closed but weak* closed! This means that you can separate $ \ell$ from $ M$ by using finitely many elements of $ X$. Should I continue?

So there are $ x_1,...,x_n$ in $ X$, complex $ c_1,...,c_n$ and $ \epsilon_j$ such that for each $ m \in M$ we have $ |m(x_j) \minus{} c_j| \geq \epsilon_j$ for some $ j$ while $ |\ell(x_j) \minus{} c_j| < \epsilon_j$ for all $ j$.

I don't get it. So, yes please

OK, the next sentence is:

If we consider the vector $ v\equal{}[\ell(x_1),\dots,\ell(x_n)]\in \mathbb C^n$ and the subspace $ L\equal{}\{[m(x_1),\dots,m(x_n)]: m\in M\}$ of the same $ \mathbb C^n$, then $ v\notin L$.

See if you can finish from here (just one step is left :)).

Is this what you had in mind? We can find a hyperplane that separates $ L$ and $ v$ which means there are coefficients so that

$ m(a_1x_1 \plus{} \cdots \plus{} a_nx_n) < \ell(a_1x_1 \plus{} \cdots \plus{} a_nx_n) \equal{} 1$

Since the $ LHS$ is a linear functional of $ m$ it has to be constantly zero. Hence $ a_1x_1 \plus{} \cdot \plus{} a_nx_n$ is annihilated by all $ m$, but not $ \ell$.

So, similar method as in the earlier post you helped me in. I don't really understand the idea, I have to admit.

Yes. Another way to present it is to say that if the vector does not belong to the subspace, there is a linear functional (in $ \mathbb C^n$) vanishing on the subspace that is not zero on that vector.

As to the idea, the only thing to understand is that, when considering weak or weak* topology, you can immediately reduce everything to a finite-dimensional problem and then just use the standard linear algebra argument.


algebra
polynomial
vector
limit
linear algebra

$ n \in \mathbb{N}$

$ B_{n,k} \equal{} C_{n}^{k}X^k(1 \minus{} X)^{n \minus{} k}$

prove that $ (B_{n,0},B_{n,1},....,B_{n,n})$ is a basis of $ \mathbb{R}_n[X]$

$ C_{n}^{k} \equal{} \frac {n!}{k!(n \minus{} k)!}$

I found this statement with a false proof I think, I'm I right?:

we only have to prove that the vectors are linearly independent

we suppose $ \sum_{k \equal{} 0}^{n} \alpha_{k} B_{n,k}(x) \equal{} 0$ for all $ x \in \mathbb{R}$
for $ x \equal{} 0$ and $ x \equal{} 1$ we have immediately $ \alpha_0 \equal{} \alpha_n \equal{} 0$ O.K
then:

$ \alpha_1 n x(1 \minus{} x)^{n \minus{} 1} \plus{} x.P(x) \equal{} 0$ for all $ x \in \mathbb{R}$  P=... some polynomial with $ P(0)\equal{}0$

we simplify by $ x$ then
$ \alpha_1 n (1 \minus{} x)^{n \minus{} 1} \plus{} P(x) \equal{} 0$(*)  [u]for all [/u]$ x \in \mathbb{R}$

then for $ x \equal{} 0$ we obtain $ \alpha_1 \equal{} 0$

the error is that (*) is true for $ x$ only in $ \mathbb{R^*}$

If a polynomial is zero for all $ x\in\mathbb{R}^*$, can it be nonzero at $ 0$ ?

  darij

Helo

$ \bullet$$ \bullet$  [b]FOURRIER[/b] :
 there is  not error because :
since  you  have $ (\forall x \in {\mathbb R}^*) \quad \alpha_{1}n (1 \minus{} x)^{n \minus{} 1} \plus{} P(x) \equal{} 0$
you ca say :  $ \lim_{x \to 0} (\alpha_{1}n (1 \minus{} x)^{n \minus{} 1} \plus{} P(x)) \equal{} 0$ ...

$ \bullet$ $ \bullet$ for  your question :
since  $ B_{n,k} \in {\mathbb R}_n[X]$  for  each  $ k \equal{} 0,...,n$  and $ n \plus{} 1$  is the dimension
of $ {\mathbb R}_n[X]$ it suffice to  prove that  the  family  is  free.It's  easy to do  this since
$ B_{n,0}(0) \neq 0$   and  $ B_{n,k} (0) \equal{} 0$  for  $ k > 0$....

$ \bullet$ $ \bullet$  in  general  if   $ (P_k)_{k \in I}$    is  a  family  of  polynoms  such  that $ P_k(0) \neq 0$  for each $ k \in I$ ($ I \subset {\mathbb N}$  and  $ I \neq \emptyset$)
we can  prove  that  the  family $ (X^k P_k)_{k \in I}$  is  free   in  $ {\mathbb R}[X]$

$ \bullet$ $ \bullet$  for the generalisation above let $ n \in {\mathbb N}$  and scalars $ a_k$  , $ k \in I_n \equal{} \{0,...,n\}$ such that : $ \sum_{k \equal{} 0}^n a_k X^k P_k \equal{} 0$
$ \bullet$ if $ n \equal{} 0$  it is  clair
$ \bullet$ if $ n > 0$,we  can  use  an inductive reasonning on the set $ I_n$ to prove that al scalars $ a_k$ are zero.
in fact :putting forall $ k \in I_n \quad Q_k \equal{} X^k P_k$  since $ Q_0(0) \neq 0$  and  $ Q_k(0) \equal{} 0$ for all $ k > 0$ , we have $ a_0 \equal{} 0$.
Suppose for $ m \in\{0,...,n \minus{} 1\}$ that : $ a_0 \equal{} ... \equal{} a_m \equal{} 0$   ,  then  we  have  : $ X^{m \plus{} 1}(a_{m \plus{} 1} \plus{} X R(X)) \equal{} 0$
where $ R$ is  some polynom and  since  the  ring  $ {\mathbb R}[X]$ is without zero divisors we have : $ a_{m \plus{} 1} \plus{} X R(X) \equal{} 0$. this gives  $ a_{m \plus{} 1} \equal{} 0$.

$ \bullet$ $ \bullet$  if   we  put  : $ P_k \equal{} C_n^k (1 \minus{} X)^k$  your  question is a particulare case to the above generalization


There are 51 gold medals and 50 silver medals, let g1<g2<g3......<g51 and s1<s2<s3......<s50 be the weights of the medals. One Know the order of the gold medals weights and the order of the silver medal weights But one has to find the order of all weights and we are allowed to ask only questions of the type:            is gi<or >sj? for some i and j. determine the smallest neccesary number of questions to determine the order of all weights.

I think the weights are distinct, and, if I remember correctly, we had to find the median wieght, not the order of all the weights. :?


Given real numbers $ x,y,z\geq \minus{}1$ satisfying $ x^3 \plus{}y^3 \plus{} z^3 \geq x^2 \plus{} y^2 \plus{} z^2$, prove that $ x^5 \plus{} y^5 \plus{} z^5 \geq x^2 \plus{} y^2 \plus{} z^2$.

[quote="arqady"][quote="hollandman"]Given real numbers $ x,y,z\geq \minus{} 1$ satisfying $ x^3 \plus{} y^3 \plus{} z^3 \geq x^2 \plus{} y^2 \plus{} z^2$, prove that $ x^5 \plus{} y^5 \plus{} z^5 \geq x^2 \plus{} y^2 \plus{} z^2$.[/quote]
If $ \{x,y,z\}\subset[ \minus{} 2, \plus{} \infty)$ and $ x^3 \plus{} y^3 \plus{} z^3 \geq x^2 \plus{} y^2 \plus{} z^2$ then
$ x^5 \plus{} y^5 \plus{} z^5 \geq x^2 \plus{} y^2 \plus{} z^2$ is still true. :wink:[/quote]
We have
\[ x^5\minus{}x^2\plus{}3(x^2\minus{}x^3)\equal{}x^2(x\minus{}1)^2(x\plus{}2) \ge 0.\]
Applying similar to $ y,z$ and adding them up, we get
\[ (x^5\plus{}y^5\plus{}z^5\minus{}x^2\minus{}y^2\minus{}z^2)\plus{}3(x^2\plus{}y^2\plus{}z^2\minus{}x^3\minus{}y^3\minus{}z^3) \ge 0\]
But $ x^2\plus{}y^2\plus{}z^2\minus{}x^3\minus{}y^3\minus{}z^3 \le 0$, hence
\[ x^5\plus{}y^5\plus{}z^5\minus{}x^2\minus{}y^2\minus{}z^2 \ge 0\]
Done!
:)


Let $ d$ be any positive integer whose units digit is equal to $ 1,3,7$ or $ 9$.
Show that $ d$ has a multiple all of whose digits are $ 1$

[hide]For any positive integer $ d \equal{} 10q \plus{} m$ with units digit $ m\in\{1,3,7,9\}$ we have $ gcd(10,10q \plus{} m) \equal{} 1$. Therefore there exists some positive integer, $ n_0$, such that $ 10^{n_0}\equiv1\pmod{d}$.  Now we have $ d\mid10^{n_0} \minus{} 1$.  There are two cases:

     $ (1) \,9\nmid d \Rightarrow d\mid\frac {10^{n_0} \minus{} 1}{9}$ and we are done.

     $ (2) \,9\mid d \Rightarrow \frac {d}{9} \equal{} x$ where $ x \equal{} 10q_0 \plus{} m_0$.  We know that $ x$ has at most one less factor of $ 9$ than $ 10^{n_0} \minus{} 1$, so $ x\mid \frac {10^{n_0} \minus{} 1}{9}$, and we must show that $ d \equal{} 9x\mid \sum_{i \equal{} 0}^k10^i$ for some positive integer $ k$. We have $ xq_1 \equal{} \sum_{i \equal{} 0}^l10^i$ for some $ l$ and $ q_1$, and we can construct a positive integer $ v \equal{} \sum_{i \equal{} 0}^810^{i(l \plus{} 1)}$.  Note that $ 9\mid v$.  Therefore we have $ d \equal{} 9x\mid vxq_1 \equal{} \sum_{i \equal{} 0}^{9l \plus{} 8}10^i$. $ \Box$
  [/hide]

[hide]Another way to look at it:
have the sequence of  numbers
1
11
111
1111
11111....

divide each one  of these by d, and notice  the remainder..
PigonHole tells us , since d is finite, there must be at least two above numbers which will give the same remainder...

Let these be,
111... (m 1's) ), and 111...(n  1's), subtract smaller from the larger, and you have 111...000  which is divisible by d, remove 0's in the end (as d and 10 have no common factors) and you are done.[/hide]

Really nice solution, [b]Gyan[/b]  :)
I never thought such a simple yet elegant solution existed. Thanks.


[hide]There are two cases: no boys are on the committee or one boy is on the committee. For the first case, there are $\binom{4}{3}$ different committees. There are $\binom{3}{1}\cdot\binom{4}{2}$ different committees in the second case.
So, we evaluate
$\frac{\binom{4}{3}+\binom{3}{1}\cdot\binom{4}{2}}{\binom{7}{3}}$, which gives us the answer $\frac{22}{35}$.[/hide]

[quote="rcv"]From a group of 3 boys and 4 girls, a committee of 3 is to be picked at random.  What is the probability that there will be more girls than boys on the committee?  Express your answer as a common fraction.[/quote]

Isn't this a little too easy for it to be National Team #10?
[hide]TOTAL: $\binom{7}{3}= 35$

We can have 2 girls and 1 boy, or all girls.

For the former, there are $\binom{4}{2}\cdot \binom{3}{1}= 18$ ways.

For the latter, there are $\binom{4}{3}= 4$ ways.

TOTAL: $22$ ways

So it is $\frac{22}{35}$.[/hide]


Hey guys i was just wondering if anyone is willing to put up the answers to the fermat contest because i am wanting to know how i did

My answers are not official,
1.E
2.B
3.E
4.A
5.C
6.A
7.B
8.B
9.E
10.C
11.E
12.A
13.B
14.C
15.C
16.D
17.B
18.E
19.D
20.E
21.D
22.C
23.A
24.C
25.B

Hoping there are no stupid mistakes......

[quote="Iam1337"]Well i didnt get that last two questions so nothing to say there however i disagree with 23.

I ending up getting E)58 anyone disagree?[/quote]

Nope, it's A


[color=green]dear friend resolve this exercice[/color]

$f'(x)=2x-2m=0\Leftrightarrow x=m$.

We consider two cases:

Case 1: $m\not\in[0,1]$:
So the minimum of the quadratic is outside the given range, so it's minimum and maximum on $[0,1]$ will be $f(0)$ and $f(1)$, so it is sufficient to find conditions for $|f(0)|\le1$ and $|f(1)|\le1$. $|f(0)|=\left|-m^2-\frac{3}{4}\right|=m^2+\frac{3}{4}\le1\Leftrightarrow |m|\le\frac{1}{2}$. So $-\frac{1}{2}\le m<0$.  It is easy to confirm that this condition is also sufficient for $f(1)\le1$.

Case 2: $m\in[0,1]$:
In this case, the minimum of $f$ will be at $f(m)$, and the maximum (by convexity) will be at $f(0)$ or $f(1)$.
$|f(m)|=\left|-2m^2-\frac{3}{4}\right|=2m^2+\frac{3}{4}\le1\Leftrightarrow|m|\le\frac{1}{2\sqrt{2}}$, so we must have $0\le m\le\frac{1}{2\sqrt{2}}$.  Again, it is easy to confirm that this condition is sufficient to also satisfy $|f(0)|\le1$ and $|f(1)|\le1$.

Combining the two cases, we get $-\frac{1}{2}\le m\le\frac{1}{2\sqrt{2}}$.


topology
function
geometry
3D geometry
sphere
inequalities
vector

How can I prove that the set of bounded continuous mappings $\mathcal{C}(X,\mathbb{R})$, where $X=\{1,2,\ldots ,n\}$, equipped with the discrete metric and $\mathbb{R}^n$ with the usual Euclidean metric have the same open sets?

[quote="julien_santini"]There is an obvious isomorphism you can build up. Now all you have to do is to check that this bijection is bicontinuous.[/quote]

Isomorphism? The word has not been mentioned in this present course (on topology). Homeomorphism on the other hand, but it's mentioned for the first time in the chapter following that where I found this exercise.

I know I may represent a function as a array $x=(x_1,\ldots ,x_n)$. My idea after that was to prove the two spaces have the same open [i]spheres[/i] (balls), because as we have proved, any open set is a union of open spheres. However, I don't think they do have the same open spheres (because I couldn't prove so). On the other hand any ''usual sphere'' in $\mathbb{R}^n$ is a union of spheres in $\mathcal{C}(X,\mathbb{R})$, i.e. one-point sets. But an open sphere in the discrete metric is either just a point or he full space. Since the full space is open I guess we can say it's a union of open spheres as well, but one-point sets are not open in the usual Euclidean  norm. My idea is probably quite bad, so is there a better way to proceed? Thanks.

I completely misunderstood that the discrete metric was on $X$; while the distance between functions is measured by the supremum norm (or the metric induced from this norm).

Here's the solution I came up with. Please feel free to citisize it!

Let $f(i)=x_i$, $x=(x_1,\ldots ,x_n)$. From
\[ \max_i |x_i-y_i|\le \|x-y \|\le \sqrt{n} \max_i |x_i-y_i| \]
it follows that for any open ball in one of the metrics, there is an open ball in the other metric contained in it.

To show the two spaces have the same open sets I prove they have the same open spheres (usually called balls, but this book calls them spheres), since any open set is a union of open spheres. Let $S_1$ be an open sphere in $\mathcal{C}(X,\mathbb{R})$. Let $x\in S_1$. Then, since $S_1$ is open, there is an open sphere $S_2$ (still in $\mathcal{C}(X,\mathbb{R})$) centered at $x$ and contained in $S_1$. Within $S_1$ there is, by the above arguments, an open sphere $S_{\alpha }$, open in $\mathbb{R}^n$ and centered at $x$. It follows that $S_1$ is the union of the $S_{\alpha }$:s and thus that $S_1$ is open in $\mathbb{R}^n$. The other way is similar.

Is there something wrong with this? Thanks.

Yeah that's the idea ... be careful about the terminology though ... the two sets do not have the same open sets, since we are not talking about the same objects. However, via the obvious identification: $\phi: f \rightarrow (f(1),\dots,f(n))$, we obtain an homeomorphism (in order to show that, you can use the inequalities you gave in your previous post, and the same idea).

Notice that this homeomorphism is also a linear isomorphism of normed vector spaces. Since they are finite-dimensional, this isomorphism is also a homeomorphism (this is more general: if $E$ is a $\mathbb{C}$ (or $\mathbb{R}$) - topological vector space linearly isomorphic to $\mathbb{C}^{n}$ (or $\mathbb{R}^{n}$), then any isomorphism between the $2$ is an homeomorphism).


geometry
3D geometry
number theory
prime numbers
modular arithmetic

Find all prime numbers $ p$ such that $ 32p\plus{}1$ is the cube of an integer.

[hide="Solution"] $ 32p \equal{} (n\minus{}1)(n^2 \plus{} n \plus{} 1)$.  Since the right factor is odd, $ 32 | n\minus{}1$.  Write $ n \equal{} 32k \plus{} 1$ and

$ p \equal{} k((32k\plus{}1)^2 \plus{} (32k\plus{}1) \plus{} 1)$.

Since the right factor cannot equal $ 1$, we have $ k \equal{} 1, p \equal{} 33^2 \plus{} 33 \plus{} 1 \equal{} 35937$.  Clearly $ 2, 3, 5, 11 \not | p$.  Moreover,

$ p \equiv (\minus{}2)^2 \plus{} (\minus{}2) \plus{} 1 \equiv 3 \bmod 7$
$ p \equiv (\minus{}6)^2 \plus{} (\minus{}6) \plus{} 1 \equiv 31 \bmod 13$
$ p \equiv (\minus{}1)^2 \plus{} (\minus{}1) \plus{} 1 \equiv 1 \bmod 17$
$ p \equiv (\minus{}5)^2 \plus{} (\minus{}5) \plus{} 1 \equiv 21 \bmod 19$
$ p \equiv 10^2 \plus{} 10 \plus{} 1 \equiv 111 \bmod 23$
$ p \equiv 4^2 \plus{} 4 \plus{} 1 \equiv 21 \bmod 29$
$ p \equiv 2^2 \plus{} 2 \plus{} 1 \equiv 7 \bmod 31$

so $ p$ is indeed prime. [/hide]

[hide="solution"]$ 32p \plus{} 1 \equal{} x^3\Rightarrow 32p \equal{} x^3 \minus{} 1 \equal{} (x \minus{} 1)(x^2 \plus{} x \plus{} 1)$. Since $ x^2 \plus{} x$ is even, $ x^2 \plus{} x \plus{} 1$ is odd, and all the powers of 2 must be in $ x \minus{} 1$. Now $ p$ must go into either $ x \minus{} 1$ or $ x^2 \plus{} x \plus{} 1$, but if $ p$ goes into $ x \minus{} 1$, then $ x^2 \plus{} x \plus{} 1 \equal{} 1$ and $ x \equal{} 0$, which is impossible. So $ p \equal{} x^2 \plus{} x \plus{} 1$. $ 32 \equal{} x \minus{} 1\Rightarrow x \equal{} 33$, and $ p \equal{} 1089 \plus{} 33 \plus{} 1 \equal{} 1123$, and we can test all primes up to $ \sqrt {1123} < 34$, and we see that $ 1123$ is prime. So the only solution is $ p \equal{} 1123$.[/hide]

@t0rajir0u: how did you come up with 35937? That isn't prime anyways; 3+5+9+3+7=27.

Well, $ {33}^3\equal{}35937$, but I'm not sure why he would say that...

Sorry; I used the wrong keys on the Windows calculator.  The correct numerical value is $ p \equal{} 1123$, which (by my arguments) is prime.

Is there a proof that your argument works?

It's well-known that to test a number's "primeness", you need only check the prime numbers less than the squareroot of the number.  Thus, he proved that 1123 is not equivalent to 0 mod p for any p<33.

Yeah I know that, but how did he come up with those numbers that he squared, added itself, and then added 1?

$ p \equal{} 33^2 \plus{} 33 \plus{} 1$, so if we want to find the remainder when $ p$ is divided by a prime it suffices to find the remainders when $ 33, 33^2$ are divided by that prime.  How much modular arithmetic do you know?

ah. ok.

I know mods, addition, subtraction, multiplication, how to find the units and tens digit of 7^(7^(7^7)), and I almost have memorized Wilson's Theorem.

I suppose that this is a bit strong for this problem, but it is a nice line of logic anyway.

Suppose that $ p|x^2\plus{}x\plus{}1$. Then $ p|(x^3 \minus{} 1)$ where $ p$ is a prime. So $ p\not | x$, otherwise $ p|1$, which is nonsense. So then by fermat's little theorem, $ p|x^{p \minus{} 1} \minus{} 1$. Now by a common lemma, those two imply that $ p|x^{(3,p \minus{} 1)} \minus{} 1$.

If $ (3,p \minus{} 1) \equal{} 1$, then $ p|(x \minus{} 1)$, but then $ p|x^2\plus{}x\plus{}1\equal{}3$, so $ p\equal{}3$.

If $ (3,p \minus{} 1) \equal{} 3$, then $ p \equal{} 3k \plus{} 1$, and since $ \sqrt {1123} < 34$, we check the primes less than $ 34$:
\[ 2,3,5,7,11,13,17,19,23,29,31
\]
We can eliminate the primes that are not $ 1\mod 3$, so we must check
\[ 7,13,19,31
\]
This is easy computation because $ (a \plus{} m)^k\equiv (a)^k\mod m$ for integers $ m > 1$, $ a,m$.


Express the next term in the sequence as a decimal: $ 0, 0.5, 0.\overline{6}, 0.75 \dots .$

Just to make things clearer:

$0=\frac{0}{1}$
$0.5=\frac{1}{2}$
$0.\overline6\overline6\overline6=\frac{2}{3}$
$0.75=\frac{3}{4}$
The next term$=\frac{4}{5}$

Answer: $0.8$


geometry
3D geometry
sphere
symmetry
geometric transformation
rotation
analytic geometry

Well, it just means to divide the surface of the sphere into n parts such that each part contains exactly one given point.

should the section be connected? I think the if we have no restriction about the n parts it is not hard but if they should be connected i have no idea :(

all is required is to divide the sphere's surface into n identical parts, so i assume they don't have to connect to each other, just make sure all of them together cover the whole surface...but maybe i'm not understanding your question? :)

[quote="al.M.V."]Prove that given n arbitrary points on the surface of a sphere, one can always divide the surface into n identical sections such that each section contains exactly one given point.[/quote]

This is easy to do but hard to write without a picture.  Find a diameter of the sphere in generic position with respect to the points (it does not pass through any point, and no plane perpendicular to the diameter contains 2 of the points).  You can start from the rotationally symmetrical partition into $n$ "slices of an orange" through the diameter, and continuously twist it into a partition that has the same symmetry under $2 \pi / n$ rotation, but each piece has exactly one of the points.

I wanted to do the exact same thing when I first saw the problem, but are you positive it can always be done? Think about the following arrangement of $6$ points on the sphere: three of them are very close to a point $A$, while the other three are very close to a point $B$ obtained from $A$ through a rotation of $90^{\circ}$ around the center of the sphere performed in a plane passing through the center.

If the three points in a cluster are close enough to each other, then this would force one of the endpoints of the diameter defined by the slicing to be very close to those points too, but then this would make the other cluster be too far away from such an endpoint.

Maybe I just misunderstood your construction? :?

Yes, I think it can be done by a linear interpolation.
Think of the sphere as a globe and the diameter as North-South axis, so that we have a fixed way of measuring angles on the "latitude" circles perpendicular to the diameter.
Each point is at a different latitude.
Surround each of the $n$ point by an interval on its "circle of latitude" of angle $2 \pi /n$, with the point in the center of the interval, and rotate this interval to divide that circle into $n$ equal intervals; number the intervals clockwise $1,2,...,n$ so that the interval containing the point in its center, is numbered $1$.
Choose the path $f(y)$ from North to South, by linear interpolation of the relevant angles (linear in the $y$ coordinate measuring latitude = position on the axis of the globe),  so that when the path reaches point $i$ it is at the left endpoint of the interval numbered $i$.   Choose piece number $1$ of the partitition to have, for each $y$, the all the points at height $y$ with angle in the interval $(f(y), f(y) + 2 \pi /n)$, and the other pieces to be rotations of this one. That's all.

I think what you're saying works, fleeting_guest, as soon as you can choose a direction (your generic line does this perfectly), every point can be obtained by rotating the interval. I don't understand 100% of what you are saying (a bit deep for me... :P ) but I can get the image and I think it works. :D

I will be happy to explain any part of the argument.  By stereographic projection it is equivalent to finding a decomposition of the plane into $n$ pieces, symmetric with respect to a $2 \pi / n$ rotation around some point, with each piece having one point inside (and the same construction can be repeated).

A harder problem (at least, I see no solution other than Banach-Tarski type arguments) is to decompose the sphere into $n$ isometric pieces which form a partition in the set-theoretic sense, i.e. each point belongs to exactly one piece.


Adrian teaches a class of six pairs of twins. He wishes to set up teams for a quiz, but wants to avoid putting any pair of twins into the same team. Subject to this condition:
i) In how many ways can he split them into two teams of six?
ii) In how many ways can he split them into three teams of four?

[quote="zhaoli"][quote="shyong"]The word " two team of six " means two teams each with six people or six teams each with two people . Well , if it is the first one then it is just one way to do it  :huh: and I guess it wont be that easy  :huh:[/quote]

The first meaning should be!

In the first case, there are more than one way, it should be $2^5$ ways if the two teams are identical. It is $2^6$ if the two teams are different.[/quote]

Opps  :blush:  I somehow thought that each " twins " are actually the same person . :P

Im interested in the (ii) one , any idea or hint?

This is rather trivial if I've understood it correctly...

Let $A, B, C$ be the three teams; $x$ the number of pairs of twins with one in $A$ and one in $B$; $y$ those with one in $B$ and one in $C$; $z$ those with one in $A$ and one in $C$. Then $x+y=y+z=z+x=4$. This implies $x=y=z=2$. So we want to split the $6$ pairs into 3 distinct groups of $2$. This can be done in $\frac{6!}{2^3}=90$ ways. Now, for each pair of twins, they can be arranged in two ways, so multiply by $2^6$. We get $2^3 \times 6!$.

[quote="googleplex"]Adrian is creating teams defined by the twins themselves - not putting the twins into different pigeon-holes called A, B and C.  The previous posting overcounts the number of teams every time two sets of teams differ only in the labels A, B and C.  Each distinct grouping of chilren has been counted 3!  = 6 times.  Therefore divide the answer 2^3 x 6! by 6 to get the correct answer:  960.

Example:  Call the pairs of twins (t11, t12), (t21, t22), ..., (t61, t62).

Then the set of teams:
A = t11, t21, t31, t41      B = t12, t22, t51, t61   C = t32, t42, t52, t62
and
A = t12, t22, t51, t61      B = t32, t42, t52, t62   C = t11, t21, t31, t41
represent exactly the same groups of children.  (The second A is the first B, the second B is the first C, the second C is the first A).  Four other similar identical rearrangements exist.

If Adrian was trying to create, say, a group of 4 musicians, 4 chess players and 4 gymnasts, then the teams would all be distinct and you would not then need to divide by 3! i.e. 2^3 x 6! would be correct.[/quote]I agree


calculus
integration
real analysis
real analysis unsolved

f is continuous on [0,1].
$ \forall$x $ \in$[0,1],$ f(x)\geq0$ and $ \int_{0}^{x}f(t)dt \geq f(x)$
show that:
f(x)=0

thanks~ :lol:

Notice that $ g\left( x \right) \equal{} \frac {{\int\limits_0^x {f\left( t \right)dt} }}{{{e^x}}},x \in \left[ {0,1} \right]$
is decreasing so $ g\left( x \right) \le g\left( 0 \right) \Leftrightarrow g\left( x \right) \le 0$. Also $ g\left( x \right) \ge 0$
 since $ f\left( t \right) \ge 0$ and f is continuous. That means $ 0 \le g\left( x \right) \le 0 \Leftrightarrow g\left( x \right) \equal{} 0,\forall x \in \left[ {0,1} \right]$. Rest are easy :)


let a,b,c>0. Prove that
$ \frac{1}{{\left(a\plus{}b \right)}^{2}}\plus{}\frac{1}{{\left(b\plus{}c \right)}^{2}}\plus{}\frac{1}{{\left(c\plus{}a \right)}^{2}}\geq \frac{3\sqrt{3abc\left(a\plus{}b\plus{}c \right)}{\left(a\plus{}b\plus{}c \right)}^{2}}{4{\left(ab\plus{}bc\plus{}ca \right)}^{3}}$

[quote="nguoivn"]It's an old problem of ductrung. Indeed, your proof is not so nice with me.
See our proofs (by AM-GM) here   :maybe:
http://www.mathlinks.ro/viewtopic.php?t=266355[/quote]
oh my solution by SOS http://ddbdt.co.cc/forum/problem-day/36-inequality.html


Personally, I think people should just stop answering your questions. I'm not trying to be mean or something, but seriously.

1) You'll never learn anything.
2) You also need to seriously use LaTeX.

[quote="And 1"]1
5x - --- if 2x+3=19
        2

Simplify:

1
-----
-5^-2

6r^2st^3 - 9rs^2t^2
-------------------------
3rst[/quote]

Sorry but I don't understand this problem because you can barely read it. Try LaTeX. Click on the link on your left. That should help people understand your question a lot better.

I think the last two things are fractions.

Besides, I agree with Zenith Xan~


calculus
analytic geometry
geometry
3D geometry

Find the volume of the solid formed by rotating the triangle with vertices (-4,1), (-2,4) and (-1,1) about the line y=-3x-2.

[quote="max_tm"]hmmm, sorry to kindof hi-jack the thread but if my question is answered it would lead to solution (albeit a calculus based solution):

Is there some generalization of volume of revolution around lines other than the x or y axis? Maybe something with rotating the axes (is that the plural of axis?  :P) with polar coordinates or something?[/quote]
You don't have to do calculus to do this problem. Once you've realized that two points of the triangle are on the line then it can be easily solved.

yeah, just two cones  :lol:

i was too lazy to do it earlier and didn't see easy that the Y was 1 for two sides and the equation forr line goes thru them. lol wow im silly

im a country bumpkin


What is the biggest integer which is equal to the product of its digits?

Is there any solution other than 9.
I don't think so.
Because, for a two digit no. 99>9*9.
Similarily, 999999...9>9*9*9*9*9*9*...*9

Is there a better soln.? :roll:

All I know is that it can't contain any 1's or 0's.

Well that was not of too help,Hexaditidom. :oops:

if u say $a_0,...,a_n$ are the digits of the number, then we want to solve:

\[\sum a_i\cdot 10^i -\prod a_i =0\]

say we take instead of $a_n$, $a_n-1$ then it becomes clear that
$\sum a_i\cdot 10^i -\prod a_i> \sum a_i\cdot 10^i -\prod a_i-10^n+a_0\cdots a_{n-1} = (a_n+1)\cdot 10^n +a_{n-1}\cdot 10^{n-1}+...+a_0-(a_n-1)a_{n-1}\cdots a_0$
so $\sum a_i\cdot 10^i -\prod a_i$ is for a number like $\overline{1a_{n-1}...a_0}$ always smaller than any number with another $a_n$ (with an $n>1$) but than we can drop the $1$ in the beginning of the number and $\sum a_i\cdot 10^i -\prod a_i$ will be smaller again ($10^n$ smaller) so we can reduce it to a 2-digit number...

but for $\overline{ab}$ we know that for $a=1$ $10a+b-ab$ be the smallest. but for any $b$ it is still greater than $0$ so there is no such number other than a one-digit number, in other words $9$ is the answer


Let $ \triangle$ ABC such that $ AB \equal{} \dfrac{AC}2$ and $ B > 90^0$. Prove that $ C > \dfrac A2.$ :?:

$ a \equal{} BC, b \equal{} CA, c \equal{} AB; b \equal{} 2c.$ Bisector of $ \angle A$ cuts $ BC$ at $ D.$ $ CD \equal{} \frac {ab}{b \plus{} c} \equal{} \frac {_2}{^3}a,$ $ AD \equal{} \frac {\sqrt {bc(b \plus{} c)^2 \minus{} a^2}}{b \plus{} c} \equal{} \frac {\sqrt {2(9c^2 \minus{} a^2)}}{3},$ $ \cos B \equal{} \frac {c^2 \plus{} a^2 \minus{} b^2}{2ca} \equal{} \frac {a^2 \minus{} 3c^2}{2ca}.$
$ \angle C > \frac {_1}{^2}\angle A$ $ \Longleftrightarrow$ $ CD < AD$ $ \Longleftrightarrow$ $ 4a^2 < 2(9c^2 \minus{} a^2)$ $ \Longleftrightarrow$ $ a^2 < 3c^2$ $ \Longleftrightarrow$ $ \cos B < 0$  $ \Longleftrightarrow$ $ \angle B > 90^\circ.$

Do not double/triple... post and please, use meaningful post titles for your problems. If you wish, say "help me" in post description.


algebra
polynomial
algorithm
algebra unsolved

Is there an algorithm to express a given symmetric polynomial as a polynomial in the elementary symmetric polinomials? I know such representation always exist, but how can one find it?

Maybe the determinant at the end of the page

[url]http://mathworld.wolfram.com/SymmetricPolynomial.html[/url]


An arithmetic sequence consists of $ 200$ numbers that are each at least $ 10$ and at most $ 100$.  The sum of the numbers is $ 10{,}000$.  Let $ L$ be the [i]least[/i] possible value of the $ 50$th term and let $ G$ be the [i]greatest[/i] possible value of the $ 50$th term.  What is the value of $ G \minus{} L$?

[hide]Let the sequence be $ a_1,a_2,\ldots,a_{200}$. The average value is 10000/200=50, so we have $ a_1\plus{}a_{200}\equal{}100$. This means that $ 10\le a_1\le 90$.

We have the formula $ a_n\equal{}a_1\plus{}(n\minus{}1)d$. Plugging in $ n\equal{}200$ gives $ a_{200}\equal{}100\minus{}a_1\equal{}a_1\plus{}199d \iff d\equal{}\frac{100\minus{}2a_1}{199}$. Then $ a_{50}\equal{}a_1\plus{}49\cdot\frac{100\minus{}2a_1}{199}\equal{}\frac{4900}{199}\plus{}a_1(1\minus{}\frac{98}{199})$. This is monotonically increasing with respect to $ a_1$, so $ G\equal{}\frac{4900}{199}\plus{}90\cdot\frac{101}{199},L\equal{}\frac{4900}{199}\plus{}10\cdot\frac{101}{199}$, and $ G\minus{}L\equal{}\frac{80\cdot101}{199}\equal{}\boxed{\frac{8080}{199}}$.[/hide]


search
algorithm
logarithms
probability
number theory
prime numbers
advanced fields

I need a 400 diget prime number that is the resault of multiplying two prime numbers

[quote="fedja"]:what?: How can a prime number be a product of 2 prime numbers? Well, I suspect that the first "prime" is a misprint. Try to search the web for primality check algorithms (I wouldn't bother with something advanced, basic [url=http://planetmath.org/encyclopedia/SolovayStrassenTest.html]Solovay-Strassen[/url] should be good enough), implement long arithmetic (or use some existing package) and let you computer run for a couple of days. Eventually, you'll find two 200-digit primes, which you'll, hopefully, be able to multiply. Alternatively, go [url=http://primes.utm.edu/lists/small/small3.html]here[/url] for a few examples.[/quote]

Couple of days?

There are approximately $ \frac{n}{\ln n}$ primes below a given $ n$.

The probability of a number being prime at $ 200$ digits is approximately $ \frac{1}{460}$, not considering the basic primes $ 2,3,5$ this can be reduced fairly reasonably... so it wouldn't really take days

Yes, you're right, of course  :) I just exaggerated a bit :P.


N is partitioned into two sets A,B such A and B are disjoint (N=A \cup B,
A \cap B= \emptyset ).Prove that
for all c is positive real number ,exist {a_n} ,{b_n} are monotone increasing sequences (a_n in A , b_n in B) s.t. :
   \lim a_n/b_n = c

This is clarly not true if A or B is finite.
Does it miss others hypothesis?

Pierre.

I really am not sure of this, but I'll post it hoping that you'll see the mistakes (if there are any):

Let's assume there is a c>0 s.t. an/bn is outside [c-\varepsilon, c+\varepsilon] for all n>=n0 (we can make \varepsilon as small as we want).

This means that for all n large enough, for any an \in A we also have In=(an/(c+\varepsilon), an/(c-\varepsilon))  \cap N inside A. We can see that l(Ian) -> 00 as  n->00. This measn that the length of the sequences of consecutive naturals inside A is unbounded. Leet's assume c>1. the we can make \varepsilon small enough for c-\varepsilon to be >1.

Let's take n as large as we need it in the following part. Let's also take n s.t. an is the first of a sequence of consecutive integers \in A, which is the longest yet (let's call the interval I). This is impossible because for any an+t which belongs to this interval, In+k is included in A, and the reunion of these intervals forms an interval longer than I, but which "starts" earlier than I, which contradicts the assumption that I is the longest yet.

I think it's ok, but I need your opinions (I realize how terribly unclear this might look :().

I have had sol since few day ago
You have right ideas! :)
But not clear .I will check it. :?

Have you had the chance to think about it any further?

My questions:
  +I_n=(an/(c+ \varepsilon ), an/(c- \varepsilon ))  \cap N inside A (Why?)
  +l(Ian): (What's mean?)
  +00:is mean infinite?

The statement is clear for $c=1$.
Suppose WLOG $c>1$ and the statement is false. There are $\varepsilon>0$ s.t. $c-\varepsilon>1$ and \[[b(c+\varepsilon),b(c-\varepsilon)]\cap A=\emptyset\] for all large enough $b\in B$. Fix some such $b\in B$. Consider sequence of segments $I_{n}=[b(c-\varepsilon)^{n},b(c+\varepsilon)^{n}]$. They are all consist of elements of $B$. Surely for some $n$ we get $I_{n}\cap I_{n+1}\neq\emptyset$ (as $(c-\varepsilon)^{n+1}<(c+\varepsilon)^{n}$ for large $n$). It follows that $I_{k}$ will cover all natural numbers from this point on. Contradiction.


ratio
geometry
MATHCOUNTS
calculus
integration
analytic geometry
probability

I did this in other forum a lot and it always worked out well. :rotfl:  :rotfl:

Rule:

1. Answer problem with explanation
2. Post next problem

Remember that this forum is in level of Basic Forum.

[size=142]Problem 1[/size]

Find the smallest integer that's greater than $2^{12}$

For example, $2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$.

1.

2^12 =
4096

4096+1 = 4097

Ha got the first one
( probably think I'm really dumb for actually calculating that number.. ) he he

2.
Simplify

(x + 6)(x - 7)

You mean like $x^2 - x - 42$? If so, then

[size=150]Problem 3[/size]

What is the prime factorization of $225225$? (Without a calculator or anything)

[quote="4everwise"][b]Problem # 4!![/b]

If the ratio of 2 circles' radii are 1 : 8, what's the ratio of their areas? Do not calculate their areas with the formula (pi)r^2[/quote]

The only thing were doing different to each side is squaring it.  Thus the ratio of their areas are 1^2 : 8^2 or  [b]1:64[/b]

[b] Problem 5 [/b]

Using the non-zero digits construct 3 number in the ratio of 1: 2 : 3.

Hey Silverfalcon, you were right. THis is going to work out well.
 :play_ball:
I'll let somebody else give it a shot before I post another problem.

[quote="Silverfalcon"][quote="G-UNIT"][quote="4everwise"]3) 225225 = 5 x 5 x 3 x 3 x 1001

[b]Problem # 4!![/b]

If the ratio of 2 circles' radii are 1 : 8, what's the ratio of their areas? Do not calculate their areas with the formula (pi)r^2[/quote]

The only thing were doing different to each side is squaring it.  Thus the ratio of their areas are 1^2 : 8^2 or  [b]1:64[/b]

[b] Problem 5 [/b]

Using the non-zero digits construct 3 number in the ratio of 1: 2 : 3.[/quote]

Your problem is very vague. Do you mean such that three numbers have distinct digit?

Because one way is 111:222:333. If you mean other way such that they use all the digits other than 0, thenthis problem is not suitable for Basics Forum.

Next problem:

This was actually from Flander Math Olympiad but I think you guys can handle this.

Let $m = \frac{50^{50}}{25^{25}}$. Find $m$.

[hide="Hint"]

Remember that 50 = 2*25. So, $50^{50} = 2^{50} \cdot 25^{50}$. Then the rest is quite obvious.[/hide][/quote]

well it was suppose to be distinct but o well..

Problem 6

answer: [b]10 ^ 50[/b], i know i have to show my work but im too lazy.. ull end up with 2^ 50 * 5 ^ 50= 10 ^ 50

problem number seven is??
ill make it up i guess:

In a lockeroom, there are 100 lockers numbered 1-100.
A guy goes into a locker room and opens all the lockers.
The next day, he goes in and closes every locker that is a multiple of 2.
The day after that, he goes in and either opens or closes (opens the locker if it is closed, and closes the locker if it is already open) every locker that is a multiple of 3.
This pattern continues until he reaches every multiple of 100.
How many lockers are open at the end?

[quote="G-UNIT"][/quote][quote="Silverfalcon"][quote="G-UNIT"][quote="4everwise"]3) 225225 = 5 x 5 x 3 x 3 x 1001

[b]Problem # 4!![/b]

If the ratio of 2 circles' radii are 1 : 8, what's the ratio of their areas? Do not calculate their areas with the formula (pi)r^2[/quote]

The only thing were doing different to each side is squaring it.  Thus the ratio of their areas are 1^2 : 8^2 or  [b]1:64[/b]

[b] Problem 5 [/b]

Using the non-zero digits construct 3 number in the ratio of 1: 2 : 3.[/quote]

Your problem is very vague. Do you mean such that three numbers have distinct digit?

Because one way is 111:222:333. If you mean other way such that they use all the digits other than 0, thenthis problem is not suitable for Basics Forum.

Next problem:

This was actually from Flander Math Olympiad but I think you guys can handle this.

Let $m = \frac{50^{50}}{25^{25}}$. Find $m$.

[hide="Hint"]

Remember that 50 = 2*25. So, $50^{50} = 2^{50} \cdot 25^{50}$. Then the rest is quite obvious.[/hide][/quote][quote="G-UNIT"]

well it was suppose to be distinct but o well..

Problem 6

answer: [b]10 ^ 50[/b], i know i have to show my work but im too lazy.. ull end up with 2^ 50 * 5 ^ 50= 10 ^ 50[/quote]

I'll think of the answer to the distinct then but I'll do it tomorrow.. Anyway.. The answer is not $10^{50}$.

Hey, guys!  I think I see a few experienced problem solvers in this thread.  Let this be a forum for the beginning AMC-8 and MATHCOUNTS students!

Quoting from the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=148301#p148301]Using this Forum[/url] thread, "answer the challenge problems that are appropriate for your age and experience level only!"

If you placed in the top quartile of a large State's MATHCOUNTS competition, you probably should stop thinking of yourself as a "beginner".

And let's slow it down a little bit.  In other forums, the next question is usually posted after confirmation (from the question's author) that the previous answer was correct.  We're a couple of questions ahead, since the last answers posted for Questions 3 and 6 are incorrect, and 5 is in play.  We now have 4 open questions to juggle.  :juggle:

[quote="Hamster1800"]why was the previous answer [to #6] not correct?[/quote]
Thank you for providing the correct answer to #3.

It seems the original answer to #6 was correct.  My mistake.

[But it gave me a good excuse to use that cool juggling emoticon.]

[quote="tarquin"]It's rather funny that you just made that problem up because I've seen this question a few times already on the forums. I think it may be a little hard for the basics.

For a locker to remain open , he must open it an even number of times. Perfect squares are the only integers that have an odd number of factors. Since he skips all multiples of 1, all the lockers numbered perfect squares are opened an even number of times. From 1-100 there are 10 perfect squares so only 10 lockers will remain open.

Problem 8
Anna wants some apples and donuts. If there are 5 different types of apples and 6 different types of donuts at the store, how many ways can she buy 3 apples and 4 donuts?[/quote]

[hide]  C ( 5, 3)  *  C ( 6, 4)

C (5,3)= 5!/ 2! *3!= 10

C ( 6 ,4 ) = 6!/ 2! *4!= 15

Plugging this in 15 * 10= [b]150 combinations [/b]

  Yea, it seemed pretty odd how 10^50 could be wrong....[/hide]

I used quadratic too after getting that.

we want the positive answer so it is....
$\frac{5(\sqrt{34}-\sqrt{2})}{2}$

So fireemblem's was correct.
I see what i did....i forgot to multiply 200 by 4...  :wallbash_red:  :stretcher:

[quote="keeyounglee"]Find $(52+6\sqrt{43})^{3/2} - (52-6\sqrt{43})^{3/2}$[/quote]

[hide]Simplify the square roots: $(52+6\sqrt{43})=3+\sqrt{43}$
$(52-6\sqrt{43})=\sqrt{43}-3$
Use difference of cubes $(a-b)(a^2+ab+b^2)$ and get $828$
[/hide]

I think that was kind of off difficulty...
Next problem:
What is the difference between the greatest and least possible integer lengths x of a triangle of length 10, 15, x?

[quote="bedakilla"]I might be wrong  :?

[hide]Since its a triangle, two sides added must be greater than remaining side.
So least possible $x$ would be $6$, and the greatest possible value of $x$ would be
$24$ therefore, $24-6=18$[/hide]

Here is my problem.
Suppose you draw 9 circles in a way so they look like a triangle with 4 circles on each side, and you must write down a distinct integer in each circles from 1 to 9.
If the sum of the integers written on each vertex of the triangle is 12, what is the
sum of the integers on each side if the sum of the integers on each sides are equal?
(I hope you understood my question correctly :P)[/quote]

Interesting question  :)
[hide]I got 19. The total sum of all the numbers is 45. Subtract 12 for the ones not on the vertexes. You get 33. There are three sides to the triangle, so each side is 33/3=11. You add the two vertexes and since there are two vertexes, each vertex is an average of 4 and you get 11+4+4=19.
[/hide]
If this is right, you can use this next problem:

This is a mathcounts problem, but it shouldn't be too hard.
Nats '97 Sprint #8

One jar contains 40% purple n&n's. A second jar contains 60% purple n&n's. The second jar contains 50% as many n&n's as the first jar. When the contents of the two jars are mixed together, exactly what percent of the n&n's in the mixture are purple? Express your answer as a mixed number.

[hide]$46\frac{2}{3}\%$

the easiest way i found to do this was to use variables to represent everything, set equations, substitute stuff, and come out with one fraction. let a=#purple in first jar, b=total in first jar, c=#purple in second jar, d=total in second jar. so,
$\frac{a}{b}=\frac{2}{5}$

$\frac{c}{d}=\frac{3}{5}$

$\frac{d}{b}=\frac{1}{2}$

we're trying to find $\frac{a+c}{b+d}$
first, i substituted $a=\frac{2}{5}b$ for a. Then i substituted $b=2d$ for both b's to get

$\frac{\frac{4}{5}d+c}{3d}$

simplifying yields $\frac{7}{15}$[/hide]

NEXT: What positive integer $n$ makes the positive difference between 2.6 and $\frac{n}{13}$ as small as possible? This is supposed to be Basic Math.

[quote="fireemblem13"][quote="Ignite168"]This is for the problem that fireemblem answered. Since they want the answer as a mixed number percent, the answer is $46\frac{2}{3}\%$[/quote]

sorry  :blush:
the [Unparseable or potentially dangerous latex formula. Error 6 ] supposed to be $46\frac{2}{3}\%$.
i'll edit it.[/quote]

 :rotfl: the latex error is  $46\frac{2}{3}\%$. It sounds kinda funny.

[quote="fireemblem13"]
NEXT: What positive integer $n$ makes the positive difference between 2.6 and $\frac{n}{13}$ as small as possible? This is supposed to be Basic Math.[/quote]

[hide] Since we're trying to find when the difference is as small as possible, I just set them equal. $2.6=\frac{n}{13} \longrightarrow n=33.8$. So the answer is $n=\boxed{34}$ [/hide]


If $A$ and $B$ are fixed points on a given circle not collinear with centre $O$ of the circle, and if $XY$ is a variable diameter, find the locus of $P$ (the intersection of the line through $A$ and $X$ and the line through $B$ and $Y$).

[hide]
Let $T$ be the intersection of tangents to circle $O$ through $A$ and $B$. Then the locus of $P$ is given by the circle $\omega$ centered at $T$ passing through $A$ (and $B$). To prove this, first extend the tangent to circle $O$ through $A$ to intersect $\omega$ again at $C$, so that $\omega$ is the circumcircle of $\triangle ABC$. Since $\angle ABC=90^\circ$ and $\angle CAB=\frac{\widehat{AB}}{2}$, ${\angle ACB=90^\circ-\frac{\widehat{AB}}{2}}$.

Given any point $Q$, extend $\overline{QA}$ and $\overline{QB}$ to intersect circle $O$ again at $M$ and $N$, respectively. Then $Q$ is in the desired locus if and only if $M$ and $N$ are diametrically opposite points. We consider two cases.

[b]Case 1:[/b] $Q$ inside circle $O$. Then $\angle AQB=\frac{\widehat{MN}+\widehat{AB}}{2}$. The measure of arc $\widehat{MN}$ is equal to 180 degrees if and only if $\angle AQB=90^\circ+\frac{\widehat{AB}}{2}$, but for points inside circle $O$, this condition is met if and only if $CAQB$ is cyclic.

[b]Case 2:[/b] $Q$ outside circle $O$. In this case, $\angle AQB=\frac{\widehat{MN}-\widehat{AB}}{2}$, so $\widehat{MN}=180^\circ$ if and only if $\angle AQB=90^\circ-\frac{\widehat{AB}}{2}=\angle ACB$, which occurs exactly when quadrilateral $AQCB$ is cyclic.

In each case, $Q$ is in the desired locus if and only if it lies on $\omega$, making $\omega$ the locus.
[/hide]


floor function
real analysis
real analysis unsolved

Is there a set o real numbers $A$ such that the greatest lower bound of the set of those $a>0$ such that $A+a=R-A$ is 0? Here $R-A$ is the complementary set of $A$.

Let $\mathcal B$ be a basis of $\mathbb R$ over $\mathbb Q$, chosen such that it contains arbitrarily small positive reals.

If $x\in\mathbb R$ can be written as $\sum_{b\in\mathcal B}a_{xb}b$ (the sum is finite, of course), then put $x$ in $A$ if $\sum_b\lfloor a_{xb}\rfloor$ is even, and in $\mathbb R\setminus A$ if that sum is odd.


If $ \frac{1}{x} \plus{} \frac{1}{y} \equal{} \frac{1}{z}$, find the value of $ y$ when $ x \equal{} 25$ and $ z \equal{} 24$.

We know that $ \frac{1}{y} \equal{} \frac{1}{24}\minus{}\frac{1}{25}$, simplifying, we get $ \frac{1}{y} \equal{} \frac{1}{600}$. Thus $ \boxed{y \equal{} 600}$.


Given $ a, b, c > 0$. Prove that
\[ \frac{2\sqrt{3}(a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca}\plus{}\frac{a\plus{}2b}{\sqrt{a^2\plus{}2b^2}}\plus{}\frac{b\plus{}2c}{\sqrt{b^2\plus{}2c^2}}\plus{}\frac{c\plus{}2a}{\sqrt{c^2\plus{}2a^2}} \ge 5\sqrt{3}\]
 :)

From
$ \sum_{cyc} \frac{(a\plus{}2b)(\sqrt{a^2\plus{}2b^2})}{a^2\plus{}2b^2}\geq \sum_{cyc} \frac{(a\plus{}2b)^2}{\sqrt{3}(a^2\plus{}2b^2)}\geq \frac{\sqrt{3} (a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2}$
It suffices to show that
$ \frac{2\sqrt{3} (a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca}\plus{}\frac{\sqrt{3}(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2}\geq 5\sqrt{3}$
But from AM-GM
$ 2\frac{(a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca}\plus{}3\frac{(a\plus{}b\plus{}c)^2}{3(a^2\plus{}b^2\plus{}c^2)}\geq 5\sqrt[5]{\frac{(a\plus{}b\plus{}c)^6}{27(ab\plus{}bc\plus{}ca)^2(a^2\plus{}b^2\plus{}c^2)}}$
Thus it remains to show that
$ (a\plus{}b\plus{}c)^6\geq 27(ab\plus{}bc\plus{}ca)^2(a^2\plus{}b^2\plus{}c^2)$
or equivalently
$ p^6\plus{}27q^3\plus{}27q^3\geq 27p^2q^2$
which is clearly true by AM-GM
 :lol: Nice indeed, nguoivn

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that
\[ \frac {2\sqrt {3}(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} \frac {a \plus{} 2b}{\sqrt {a^2 \plus{} 2b^2}} \plus{} \frac {b \plus{} 2c}{\sqrt {b^2 \plus{} 2c^2}} \plus{} \frac {c \plus{} 2a}{\sqrt {c^2 \plus{} 2a^2}} \ge 5\sqrt {3}\]
:)[/quote]
Actually, the stronger inequality holds:
$ \frac {4(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{}\sqrt{ 3}\sum \frac {a \plus{} 2b}{\sqrt {a^2 \plus{} 2b^2}} \ge 13.$

[quote="Rose_joker"]From
$ \sum_{cyc} \frac {(a \plus{} 2b)(\sqrt {a^2 \plus{} 2b^2})}{a^2 \plus{} 2b^2}\geq \sum_{cyc} \frac {(a \plus{} 2b)^2}{\sqrt {3}(a^2 \plus{} 2b^2)}\geq \frac {\sqrt {3} (a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2}$
It suffices to show that
$ \frac {2\sqrt {3} (a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} \frac {\sqrt {3}(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2}\geq 5\sqrt {3}$
[/quote]
It can be rewrite to $ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\plus{}\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2} \ge 2$(obviously trues by Am-Gm).
Thank you very much for your care, Rose_joker  :)

[quote="can_hang2007"][quote="nguoivn"]Given $ a, b, c > 0$. Prove that
\[ \frac {2\sqrt {3}(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} \frac {a \plus{} 2b}{\sqrt {a^2 \plus{} 2b^2}} \plus{} \frac {b \plus{} 2c}{\sqrt {b^2 \plus{} 2c^2}} \plus{} \frac {c \plus{} 2a}{\sqrt {c^2 \plus{} 2a^2}} \ge 5\sqrt {3}\]
:)[/quote]
Actually, the stronger inequality holds:
$ \frac {4(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} \sqrt { 3}\sum \frac {a \plus{} 2b}{\sqrt {a^2 \plus{} 2b^2}} \ge 13.$[/quote]
Can you post your proof for this stronger ? Thank you very much, my brother  :)

[quote="can_hang2007"][quote="nguoivn"]Given $ a, b, c > 0$. Prove that
\[ \frac {2\sqrt {3}(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} \frac {a \plus{} 2b}{\sqrt {a^2 \plus{} 2b^2}} \plus{} \frac {b \plus{} 2c}{\sqrt {b^2 \plus{} 2c^2}} \plus{} \frac {c \plus{} 2a}{\sqrt {c^2 \plus{} 2a^2}} \ge 5\sqrt {3}\]
:)[/quote]
Actually, the stronger inequality holds:
$ \frac {4(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} \sqrt { 3}\sum \frac {a \plus{} 2b}{\sqrt {a^2 \plus{} 2b^2}} \ge 13.$[/quote]
See here: http://canhang2007.wordpress.com/2009/12/03/inequality-88-v-q-b-can/


Did anybody see "Cold Pizza" on ESPN2, the morning of June 3?

With Gregory and Jennifer awaiting their interview, the anchor jokingly went to a commercial with a remark about "Revenge of the Nerds".

When they came back from break, I thought both Gregory and Jennifer handled the interview very well.

Thoughts or comments?

I really didn't like the revenge of the nerds very much either but I wasn't going to complain on live national TV.
The interviewer also asked me and Greg some weird questions.

Examples:
"What did the president say when you gave him the plaque?"
"Why was this years mathcounts competition hard?"

Aren't we going to have interesting answers for these questions. *sarcasm*


combinatorial geometry
combinatorics proposed
combinatorics

We have finite white and finite black points that for each 4 oints there is a line that white points and black points are at different sides of this line.Prove there is a line that all white points and black points are at different side of this line.

We assume that there are at least 4 points otherwise the result may not hold  ;)
Moreover, we assume that both colors are used otherwise the result is obvious.... :D

Ok, now :

Lemma 1 :
There do not exist a black point which belongs to a segment whose both endpoints are white.

Proof :
Clearly, if B is black and belongs to the segment $[W,W']$, where $W,W'$ are white, then considering any other fourth point, a lin separating the black points from the white ones must meet the line $WW'$ between W and B, and between W' and B, which is impossible.

Lemma 2 :
Let $C_w$ be the convex hull of the white points. Then, there is no black point on the sides or in the interior of $C_w$.

Proof.
Since the number of white points is finite, $C_w$ is a convex polygon (possibily degenerated).
From lemma 1, we know that there is no black point on the sides of $C_w$.
Now assume that the black point $B$ is interior to $C_w$. Thus, $C_w$ is not  contained in a line. Using a triangulation of $C_w$ by the diagonals from one of its vertices, we deduce that $B$ is an interior point of the triangle $WW'W''$ whose vertices are white points.
Using this set of 4 points, any line which separate $B$ from the three vertices must meet two of the triangle sides, say $WW'$ and $WW''$. In that case, $W,W'$ are not on the same side of the line. A contradiction.

We have a similar result of the convex hull $C_b$ of the black points.

It follows that $C_w$ and $C_b$ are two disjoint convex polygons. It is now clear that we may find a separating line for these two polygons, and we are done.

Pierre.


trigonometry
limit
logarithms
real analysis
real analysis unsolved

$\forall a \in R$,Prove:
\[ \lim_{n \to \infty} \prod_{k=1}^{n+1}\cos \left(\frac{\sqrt{2k-1}}n a^2 \right)=e^{-\frac{a^4}2}  \]

ok,here is my approach.
[hide]Proof:the desired limit is equal to.
\[ \lim_{n \to \infty} \prod_{k=1}^{n+1} \ln{\cos \left(\frac{\sqrt{2k-1}}n a^2 \right)}=-\frac{a^4}2  \]
but we have:\[ \cos \left(\frac{\sqrt{2k-1}}n a^2 \right)=1-\frac{(2k-1)a^4}{2n^2}+o(\frac 1 n), n \to \infty  \]
thus:\[ \ln{\cos \left(\frac{\sqrt{2k-1}}n a^2 \right)}=-\frac{(2k-1)a^4}{2n^2}+o(\frac 1 n) ,n \to \infty  \]
Then we are done[/hide]
I worder if this is a proof?

Yes. Your proof works. You can also try a proof by corseting (first few coeff from Taylor series).
\[ x \ge \ln(1+x)\ge x-\frac{x^2} 2 \]
\[ 1-\frac{x^2} 2\le \cos(x)\le 1-\frac{x^2} 2+\frac{x^4}{4!} \]
Combining these two aproximations you will get
\[ -\frac{x^2} 2 -\frac{x^4} 8 \le \ln(\cos(x))\le -\frac{x^2} 2+\frac{x^4}{4!}. \]
using this aproximation you will get the desired limit.


In a school there $b$ teachers and $c$ students. Suppose that
a) each teacher teaches exactly $k$ students, and
b)for any two (distinct) students , exactly $h$ teachers teach both of them.
Prove that $\frac{b}{h}=\frac{c(c-1)}{k(k-1)}$.

[hide="Solution"]We'll show that the following is true:
\[b{k \choose 2}= h{c \choose 2}\].
Each side of the equation counts all the distinct sets of one teacher and two different students that he/she teaches. In the left-hand side, for each of the $b$ teachers, they can be grouped with all possible 2-subsets of the $k$ students they teach. In the right hand-side, all pairs of students amongst all $c$ students can be grouped with any of the $h$ teachers that teach both of them. The two sides are equal since we are counting the same entity. Rearranging gives the desired $\frac{b}{h}= \frac{c(c-1)}{k(k-1)}$.[/hide]

This is exactly one of the (necessary) conditions for the existence of a BIBD (Balanced Incomplete Blocks Design), gotten through typical double-counting.


geometry
3D geometry
prism
pyramid
trapezoid
trigonometry
circumcircle

Each cell of a beehive is constructed from a right regular 6-angled prism, open at the bottom and closed on the top by a regular 3-sided pyramidical mantle. The edges of this pyramid are connected to three of the rising edges of the prism and its apex $T$ is on the perpendicular line through the center $O$ of the base of the prism (see figure). Let $s$ denote the side of the base, $h$ the height of the cell and $\theta$ the angle between the line $TO$ and $TV$.

(a) Prove that the surface of the cell consists of 6 congruent trapezoids and 3 congruent rhombi.
(b) the total surface area of the cell is given by the formula $6sh - \dfrac{9s^2}{2\tan\theta} + \dfrac{s^2 3\sqrt{3}}{2\sin\theta}$

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=286[/img]

The circumradius of both the regular hexagon ABCDEF and of the equilateral triangles $\triangle UVY, \triangle VXZ$ is equal to s. The sides UW, VX of these equilateral triangles are equal to $UW = WY = YU = VX = XZ = ZV = s \sqrt 3$. UW is the horizontal diagonal of the quadrilateral TUVW. If we use vertical parallel projection to project the quadrilateral TUVW into the plane of the hexagon ABCDEF, we see that the diagonals cut each other in half (the intersection of the triangle altitude with the circumcircle is the reflection of the orthocenter in the triangle side for any triangle), hence, the quadrilateral TUVW is a parallelogram. Since the diagonals are perpendicular to each other, this parallelogram is a rhombus. The 3 rhombi TUVW, TWXY, TYZU are congruent because of the 3-fold rotational symmetry. The trapezoids ABVU, CBVW are congruent, because UV = WV (TUVW is a rhombus) and the remaining 2 pairs of the trapezoids are congruent with the first pair because of the 3-fold rotational symmetry.

Let $P \equiv TV \cap UW$ be the intersection of the diagonals of the rhombus TUVW and Q the circumcenter identical with the orthocenetr of the equilateral triangle $\triangle UWY$ (the intersection of the plane of this triangle with the 3-fold rotational axis of the structure). From the right angle triangle $\triangle TPQ$ with the side $PQ = \frac s 2$ and the angles $\angle PQT = 90^\circ,\ \angle PTQ = \theta$, we get $TQ = \dfrac{PQ}{\tan \theta} = \dfrac{s}{2 \tan \theta}$. This gives us the longer base $h_1$ of the congruent trapezoids:

$h_1 = AU = CV = EY = OQ = TO - TQ = h - \dfrac{s}{2 \tan \theta}$

Since the perpendicular rhombus diagonals cut each other in half, the shorter base $h_2$ of these trapezoids is obviously equal to

$h_2 = BV = DX = FZ = TO - 2\ TQ = h - \dfrac{s}{\tan \theta}$

The area of the trapezoid ABVU is equal to

$S_T = \dfrac{h_1 + h_2}{2} \cdot s = \left(h - \dfrac{3s}{4 \tan \theta}\right) \cdot s = sh - \dfrac{3s^2}{4 \tan \theta}$

The hypotenuse TP of the right angle triangle $\triangle TPQ$ is equal to $TP = \dfrac{PQ}{\sin \theta} = \dfrac{s}{2 \sin \theta}$. Thus we obtained the diagonals of the rhombus TUVW:

$d_1 = UW = s \sqrt 3$ (the side of the equilateral triangle $\triangle UWY$ with the circumradius s)

$d_2 = TV = 2\ TP = \dfrac{s}{\sin \theta}$

The area of the rhombus TUVW is equal to

$S_R = \dfrac{d_1 \cdot d_2}{2} = \dfrac{s^2 \sqrt 3}{2 \sin \theta}$

The area of the whole structure (6 trapezoids plus 3 rhombi) is then

$S = 6S_T + 3S_R = 6sh - \dfrac{9s^2}{2 \tan \theta} + \dfrac{3s^2 \sqrt 3}{2 \sin \theta}$

The expression contains $\sqrt 3$, not $\sqrt 2$.

[quote="yetti"]The expression contains $\sqrt 3$, not $\sqrt 2$.[/quote]Indeed. :)

Did you enjoy the problem? :roll:  :D

It was too easy for MO, even for the 9th grade MO. Most problems of the Flanders MO appear to be too easy, just computational.

Yetti

Many (not to say most) are indeed computational. But this was one of the worst ever. (except for the beginning years)
[apparently the original question was even to find a maximal surface of this :?]

Each cell of a beehive is constructed from a right regular 6-angled prism, open at the bottom and closed on the top by a regular 3-sided pyramidical mantle. The edges of this pyramid are connected to three of the rising edges of the prism and its apex $T$ is on the perpendicular line through the center $O$ of the base of the prism (see figure). Let $s$ denote the side of the base, $h$ the height of the cell and $\theta$ the angle between the line $TO$ and $TV$.
(a) Prove that the surface of the cell consists of 6 congruent trapezoids and 3 congruent rhombi.
(b) the total surface area of the cell is given by the formula $6sh - \dfrac{9s^2}{2\tan\theta} + \dfrac{s^2 3\sqrt{3}}{2\sin\theta}$
[img]https://3.bp.blogspot.com/-ClTmCPYxbyw/XWur8lMf0zI/AAAAAAAAKoY/Xpkj4HewB0kFTUdKmV3jPH420XU-UJqWwCK4BGAYYCw/s1600/2004%2Bflanders%2Bp4.png[/img]


geometry
3D geometry
sphere
function
rotation
calculus

I should be able to do this problem but for some reason, my brain's freezing on it.  Please help with the following:

A cone circumscribes a sphere with radius 1.

a. Express the volume V of the cone as a function of the base radius r.

b. Use a computer or graphing calculator to find the radius and height of the cone having minimum volume.

I only need help with part a though, that's all I need.

[color=cyan]Try looking at a planar cross section of the cone/sphere which passes through the axis of the cone.[/color]

do this wil a circle and two tangent, intersecting lines in 2d, and the use shell method to rotate and find volume.

[color=cyan]No need for shell method or other calculus -- just switch to 2D in order to get the height, then use normal formulae.[/color]


Hi guys, i really need ur help here. This is about random variables.

Suppose a group of 100 men and 100 women. They r divided randomly into 100 couples. Define the random variable X. Calculate E[X]?

thx in advance.

[quote="soccerstar123"]it seems as thouht you e new menber since you only posted 2 times.. but is this a homework problem or just a ramdon problem??

becuase it's seem farmilar...

-soccerstar123[/quote]

Yeah, i am a new user. i registered before 2 days. well, it's a homework that i have and i need help. sorry if the title was so misleading. hope anyone can help me.

thx in advance.


least common multiple
number theory proposed
number theory

You can't choose $x$ - it varies and assumes natural values but the congruence $x^{d}\equiv 1 \mod n$ remains satisfied.

[quote="Xixas"]Prove or disprove the following:

If $x^{d}\equiv 1 \mod n\ \ \forall x\colon (x, n)=1, \ \ x, d, n\in\mathbb{N}$ then $\phi(n)|d$.[/quote]
It is wrong. Let $Z/nZ=(Z/d_{1}Z)+(Z/d_{2}Z)+...+(Z/d_{k}Z)$, were $d_{1}|d_{2}|...|d_{k}$.
For all x: (x,n)=1 we have $x^{d_{k}}=1(mod \ n), \phi(n)=d_{1}*d_{2}*...*d_{k}.$
It is true if and only if, when k=1. k=1 if and only if $n=2,4,p^{k},2p^{k}$, were p is odd prime.
For example n=12 $\phi(n)=4$. If (x,12)=1, then $x^{2}=1(mod \ 12$
$4\not |2$ :maybe:

I don't want to spoil it for others, so:
[hide]$n=15, d=4$

Let $k$ be the greatest integer such that $2^{k}|n$ and $s=0$ when $k=0$, $s=k-1$ if $k<3$ and $s=k-2$ otherwise.
In general any $d$ with $x^{d}\equiv 1 \mod n$ for all $x$ coprime to $n$ is a multiple of the least common multiple of $2^{s}$ and $\varphi(p^{k})$ where  $p^{k}$ runs through all odd prime powers dividing $n$; conversly, all those $d$ work.

Prove the last statement if you want.[/hide]


Combinatorics of words
maximization
combinatorics
Extremal combinatorics
IMO Shortlist
IMO Longlist

An alphabet consists of $n$ letters. What is the maximal length of a word if we know that any two consecutive letters $a,b$ of the word are different and that the word cannot be reduced to a word of the kind $abab$ with $a\neq b$ by removing letters.

We can show that it's $2n-1$ by induction. For $n=2$ we have the word $aba$, and it's clear that we can't continue to add letters. Assume we have shown the maximal length to be $2k-1$ for $k<n$. Now we can only work with $n>2$.

First assume that each letter appears at least $3$ times. Then from the word we can extract a subword using only the first $n-1$ letters and satisfying the condition, which has length at least $3(n-1)>2n-1$ because $n>2$, but this contradicts the induction hypothesis. There is, thus, a letter $a$ which appears at most twice. Now disregard $a$ from our word. What we get is a word using only $n-1$ letters and satisfying the condition. It must then be $\le 2(n-1)-1=2n-3$. This, together with the fact that $a$ appears at most twice, shows that the length of our word on $n$ letters is $\le 2n-3+2=2n-1$.

We can construct the example $a_1,a_2,\ldots,a_n,a_{n-1},\ldots,a_1$ in order to show that $2n-1$ is attainable.


1.) Prove that if $ a$ and $ b$ are both odd integers, then $ a^4 \plus{} b^4 \minus{} 2$ is divisible by 16.

2.) That the sum of the squares of two odd integers cannot be a perfect square.

I have a proof for number 1, just want to find a more elegant proof.

hint for #1
[hide] find $ a^4 \pmod{16}$ for odd a[/hide]
solution
[hide]
assume a is odd
$ a^4 \minus{} 1 \equal{} (a^2 \plus{} 1)(a^2 \minus{} 1) \equal{} (a^2 \plus{} 1)(a \plus{} 1)(a \minus{} 1)$
$ a^2 \plus{} 1$ has one factor of two because $ a^2$ is odd so $ a^2 \plus{} 1$ is even
$ a \plus{} 1$ and $ a \minus{} 1$ are both even, but, since they are consecutive even numbers, at least one of them will be divisible by 4, so these 2 factors contribute 3 factors of 2.
that means $ (a^2 \plus{} 1)(a \plus{} 1)(a \minus{} 1)$ has 4 factors of 2 and therefore is divisible by 16.
so $ a^4 \minus{} 1 \equiv 0 \pmod{16} \implies a^4 \equiv 1 \pmod{16}$
and the same is true for odd b, so $ a^4 \plus{} b^4 \minus{} 2 \equiv 1 \plus{} 1 \minus{} 2 \equiv 0 \pmod{16}$
[/hide]
hint for #2
[hide]
find the quadratic residues (possible remainders when perfect squares are divided by the mod base) in mod 4
then find the only possible remainder when the square of an odd number is divided by 4
[/hide]
solution
[hide]
when perfect squares are divided by 4, the only remainders possible are 0 and 1.
when squares of odd numbers are divided by 4, the only remainder possible is 1.
so if a and b are odd,
$ a^2\plus{}b^2 \equiv 2 \pmod{4}$
$ a^2\plus{}b^2$ cannot be a perfect square because when it is divided by 4, it leaves a remainder than is not possible for perfect squares
[/hide]

I like the proof in 1. Well, here's mine. It's very basic.

[hide="Hint"]Odd numbers are of the form, $ 8k \plus{} 1$ (Prove it!)[/hide]

[hide="Proof"]$ a^4 \plus{} b^4 \minus{} 2 \equal{} (a^2)^2 \plus{} (b^2)^2 \minus{} 2$
$ \equal{}(8k\plus{}1)^2 \plus{} (8r \plus{} 1)^2 \minus{} 2$ for k,r in Z
I think its obvious [/hide]