induction geometry Gauss LaTeX AMC USA(J)MO USAMO

Umm...can someone please explain this to me?   As if I was four years old please...

Gauss202 explained it very nicely in this post... check it out [url]http://artofproblemsolving.com/Forum/viewtopic.php?t=5966[/url]

Okay, mathematical induction is a method of proof.  It's probably the easiest way to prove things that have to do w/sequences and series.  I'll show you an example while explaining the steps:

Prove that the sum of the first n odd numbers is [tex]n^2[/tex].

(1) <b> Show that it is true for n=1 </b>.

The sum of the first 1 odd numbers = [tex]1^2=1[/tex].  1 is the first odd number, so it =1, so you know this is true. (duh, but it's part of the process)

Since n=1 seemed pointless, I like to try another like n=3.  I asked AoPS people before and some said it was not necessary, but I like it anyway, so here:

1+3+5 = sum of first 3 odd numbers = 9
[tex]3^2[/tex] = sum of first 3 odd numbers = 9

So it works for n=3.

(2) <b> If it is true for some integer [tex]k[/tex], then it must be true for some integer [tex](k+1)[/tex].  So, show that it is true for [tex](k+1)[/tex] assuming that it is true for [tex]k[/tex]. </b>

The sum of the first [tex]k[/tex] odd numbers is:

1 + 3 + 5 + 7 + ... + [tex](2k-1)[/tex] = [tex]k^2[/tex]

The sum of the first [tex](k+1)[/tex] odd numbers is:

1 + 3 + 5 + 7 + ... + [tex](2k-1)[/tex] + [tex](2k+1)[/tex] = [tex](k+1)^2[/tex]

We need to show that this is true, so, we want to make the LHS (left hand side) of the equation look exactly the same as the RHS of the equation.

<b> 1 + 3 + 5 + 7 + ... + [tex](2k-1)[/tex] </b> + [tex](2k+1)[/tex] = [tex](k+1)^2[/tex]

We already assumed that the bold part of the LHS = [tex]k^2[/tex], so substitute that in:

[tex]k^2[/tex]+[tex](2k+1)[/tex] = [tex](k+1)^2[/tex]

Foil out the RHS:

[tex]k^2[/tex]+[tex](2k+1)[/tex] = [tex]k^2+2k+1[/tex]

So:

[tex]k^2+2k+1 = k^2+2k+1[/tex]

Cool, huh? =)

Okay, now one thing: inductive and deductive reasoning are different.  When you reason inductivley, you have to assume some things.  When you reason deductivley, you use only stuff you already know.  So the two column proofs you do in geometry employ deductive reasoning.  (math whiz people - correct me if this is some sort of egregious error)

Alright, these AoPS threads you will find helpful:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12557
(from gauss)

? I can't find it but I know he posted it in beginner's forum.
(from Osiris)

Okay, now here are some problems to practice with.  Guys, please let SingTheSorrow do them first.  In increasing difficulty:

(1) Prove: [tex]1^3 + 2^3 + 3^3 + 4^3 + ... + n^3 = \frac{n^2(n+1)^2}{4}[/tex].

(2) Prove: [tex]\displaystyle(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})...(1+\frac{1}{n})=n+1[/tex].

(3) Prove: [tex]\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} > \sqrt{n}[/tex] for [tex]n \ge 2[/tex].

(4) Prove: [tex]n! > 2^n[/tex] for [tex]n \ge 4[/tex].

(5) Prove: [tex]\displaystyle(\frac{x}{y})^{n+1} < \displaystyle(\frac{x}{y})^{n}[/tex] for [tex]n \ge 1[/tex] and[tex]0<x<y[/tex].

(6) Prove that a factor of [tex](2^{2n-1}+3^{2n-1})[/tex] is [tex]5[/tex].

Hope this helps.

Here's another problem I though was cool when I was learning induction:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11908&start=0&postdays=0&postorder=asc&highlight=

The one thing you might find confusing about my example is if you wonder "how did I know that the [tex]k^{th}[/tex] odd number was [tex](2k-1)[/tex]?"

In problems, they might give it to you in a form where they tell the nth term to you, like in (1), (2), and (3).

Here, you can just use logic (it's pretty easy to guess that's what it is), or use arithmetic sequences:

[tex]1 + 3 + 5 + 7 +...+a_n[/tex]
[tex]a_1 = 1[/tex]
You add 2 to get the next term, so [tex]d=2[/tex].
For all arithmetic sequences: [tex]a_n = a_1 + (n-1)d[/tex].
So in this situation:
[tex]a_n = 1 + (n-1)(2) = 2n - 1[/tex].

Ask if there are questions.

Dum de dum... I didn't realize I was actually going to LEARN induction...Haha, it took me so long to learn the legendre's formula...Anyway...Here I go with the answer for number 1...I'm still confused with getting the whole "k+1" thing.  Also, I don't know how do use latex, though I just downloaded the software.  so it'll be fun.

1)  I picked the number as 5, because I don't think I could PICK the number as 1 because it's an arithmetic sequence..So:
1 :^3:  + 2 :^3: + 3 :^3: + 4 :^3: + 5 :^3: = (5 :^2:(5 + 1) :^2:)/4
Simplified:
1 + 8 + 27 + 64 + 125 = 25(36)/4
Simplified again.
225 = 225.  Which is RIGHT!  Yes. :-D

Now, here's where I'm confused.  Mucho.  This is what I did so far.
1 :^3:  + 2 :^3: + 3 :^3: + 4 :^3: + (n+1) :^3: = n :^2:(n+1) :^2:

Nevermind....I forgot that uh, you're supposed to add the 1 to places...so here is the proof.
[b]1 :^3:  + 2 :^3: + 3 :^3: + 4 :^3: +... n :^3:[/b] + (n+1) :^3: = [(n+1) :^2:(n+2) :^2:]/4
We already PROVED that bolded text is equaled to n :^2:(n+1) :^2:
Therefore:
n :^2:(n+1) :^2: + (n+1) :^3 = [(n+1) :^2:(n+2) :^2:]/4
Multiplied.  Finding the LCM.  Combining like terms and the like.  (I'm just too lazy to write all of this out right now).  We get:
n :^4: + 6n :^3: + 13n :^2: + 12n + 4 = the SAME THING.

Okay, I'm matematically drained right now.  I'll tackle some later. :-P  Thanks Julie!

Okay, you are doing a good job fellow RIer (I didn't catch your name?).

[quote]We already PROVED that bolded text is equaled to...[/quote]

Ya didn't prove it, you just are ASSUMING it.  That's the whole deal w/inductive reasoning.  You just showed it for one case of n=5, that dosent necessarily mean it is true for n=any #.

[quote]I'm still confused with getting the whole "k+1" thing.[/quote]

You have to use k to so you don't get it confused w/n.  n usually represent the term number, like you have [tex]a_1, a_2, a_3, ... a_n[/tex], so [tex]a_n[/tex] is the nth term.  k and (k+1) you plug in for n, but they are not really the same thing.  Just let n=k, then let n=(k+1).

You are supposed to be formal when writing proofs, so if this was USAMO or something I would do both n=1 and n=5, but I think what you did is good.

Also, as for LaTeX, I actually don't know how to create documents either, I just know how to post on this site w/it.  I am not great w/computers *sigh*, anyway for basic commands Rep123max suggested a good website to me:

http://www.sosmath.com/CBB/docs/howtolatex.pdf

Maybe this somewhat whimsical example will help you understand the ideas that make mathematical induction work.

There is a stair with infinitely many steps. There are no steps missing in the middle anywhere.

I want to prove that I can get to every step on the stair.

I start by stepping on the first step of the stair - that is obviously doable.

Now, suppose I've arrived at the kth step. Since there are no steps missing, obviously I can get to the (k+1)th step. (*)

Then we go back to when I am on the first step. By (*), I can get to step 2; now I'm standing on step 2. Then, by (*) again, I can get to step 3; and so on. By an infinite iteration of this argument, I can show that I can get to every single step of the stair, be it the 4th or the 972348674123rd.

That is the idea behind mathematical induction. By showing that if the hypothesis works at one value, it works at the NEXT value, and finding ONE value for which it works, we can prove that all values FOLLOWING that ONE value make the hypothesis work.

Osiris, you posted a great tutorial on mathematical induction for getting started, but I can't find it.  Would you mind posting a link?

It wasn't me, it was gauss202; but the page is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12557]here[/url].

2)
With n = 1
(1/1/1) = 2
2=2 CORRECT

With n = 4
(2)(1.5)(1/1/3)(1/1/4) = 4 + 1
5 = 5 CORRECT

[b](2)(1.5)(1/1/3)(1/1/n)[/b](1/1/(n+1) = (n+1) + 1

Assuming that the bolded part is n + 1 because of the previous two examples.  Now, let's prove that it's true  for k+1.  Ergo:

(k+1) [1/1/(k+1)] = k+2

Since 7(7/1/7) equals 49 + 1 you can um, split the multiplication up. I'm sure there is a weirdo rule for that like (a)(b) = (b)(a) ...associative property?  Well, anyway.

Multiplying you get.
k+1 + (k+1)/(k+1)

And of course, x/x equals 1 so..
k+1 +1 = k + 2.  WHICH IS CORRECT.

3)

I'm confused now with the part b of matematical induction...help?  But here's what I have:
N=2

So. 1/ sqrt (1) + 1/ :rt2: >  :rt2:

I wasn't too sure about this...but I just put it in my calculator and it verfied the statement.  Can I do something different rather than rounding :rt2: to 1.41?

Now, I'm not too sure about how do do the next part either...just assume that I showed all my work, but now how do I show sqrt (k) + 1/ sqrt (k+1) > sqrt  (k + 1)?

I attempted the others but I'm sleeping now.

You brought up a very valid point.  I am not satisfied with the answer that I come up w/for this one, and the solutions manual does not give a great answer.  I will let someone else answer this one.

#2 looks good.

3) It obviously works for n = 2.

Start with the obviously true statement [tex]2 > \sqrt{2}[/tex]. Divide 2 on both sides to get [tex]1 > {{\sqrt{2}}\over{2}}[/tex]. Now add [tex]{{\sqrt{2}}\over{2}}[/tex] to both sides to get [tex]1 + {{\sqrt{2}}\over{2}} > \sqrt{2}[/tex]. Divide the numerator and the denominator of the fraction by [tex]\sqrt{2}[/tex]. You should arrive at the desired inequality.

Suppose it works for n = k; that is,

[tex]\displaystyle {{1}\over{\sqrt{1}}} + {{1}\over{\sqrt{2}}} + \ldots + {{1}\over{\sqrt{k}}} > \sqrt{k}[/tex].

Add [tex]{{1}\over{\sqrt{k+1}}}[/tex] to both sides. We obtain a RS of [tex]\sqrt{k} + {{1}\over{\sqrt{k+1}}}[/tex]. We then wish to prove the inequality

[tex]\displaystyle \sqrt{k}+{{1}\over{\sqrt{k+1}}} > \sqrt{k+1}[/tex].

Can you prove this inequality?

[quote="SingTheSorrow"]3)

I'm confused now with the part b of matematical induction...help?  But here's what I have:
N=2

So. 1/ sqrt (1) + 1/ :rt2: >  :rt2:

I wasn't too sure about this...but I just put it in my calculator and it verfied the statement.  Can I do something different rather than rounding :rt2: to 1.41?

Now, I'm not too sure about how do do the next part either...just assume that I showed all my work, but now how do I show sqrt (1): + 1/ sqrt (k+1) > sqrt  (k + 1)?

I attempted the others but I'm sleeping now.[/quote]

Umm... if you don't want to use a calculator to prove that [tex]1+\frac{1}{\sqrt{2}}>\sqrt{2}[/tex] you can do this:
Make the left side one fraction
[tex]\frac{\sqrt{2}+1}{\sqrt{2}}>\sqrt{2}[/tex]
Assume it is true and multiple by sides by  :rt2: .

[tex]\sqrt{2}+1>2[/tex]
[edited]
[tex]\sqrt{2}>1[/tex]
The last equation is obviously true and so your assumption is true.  I think that you can do this but I am not sure.  That is the best way that I can think of.

[quote="joml88"]
Umm... if you don't want to use a calculator to prove that [tex]1+\frac{1}{\sqrt{2}}>\sqrt{2}[/tex] you can do this:
Make the left side one fraction
[tex]\frac{\sqrt{2}+1}{\sqrt{2}}>\sqrt{2}[/tex]
Assume it is true and multiple by sides by  :rt2: .

[tex]\sqrt{2}+1>1[/tex]

[tex]\sqrt{2}>0[/tex]
The last equation is obviously true and so your assumption is true.  I think that you can do this but I am not sure.  That is the best way that I can think of.[/quote]

You made a tiny mistake: :rt2: * :rt2: = 2 not 1.

That's what I was having trouble with, I got it into a bunch of different forms, and none of them shouted "I am bigger than [tex]\sqrt{k}[/tex]!"  I don't really know how to prove innequalities...Only AM-GM, if that is what you are referring to...

However this should follow w/what you did for n=2.

Start w/ the statement [tex]k>\sqrt{k}[/tex].

EDITED: Bunch of unhelpful blahblah...

Not exactly what we want...maybe SingTheSorrow can answer this...

JS1527 wrote:That's what I was having trouble with, I got it into a bunch of different forms, and none of them shouted "I am bigger than [tex]\sqrt{k}[/tex]!"

I haven't been following this topic much, but I assume you are trying to prove that k >= sqrt[k]

Solution:

[hide]

If k = 0 or 1, then there is equality.  If k is greater than 1, then

k(k-1) >=0

k^2 - k >=0

k^2 >=k

k >=sqrt[k]

QED

[/hide]

OK:

We start with

[tex]\displaystyle 1 > {{1}\over{k + 1}}[/tex]. This is true for any k :ge: 1. Add k to both sides, and move 1 to the right:

[tex]\displaystyle k > k - 1 + {{1}\over{k + 1}}[/tex]. Rewrite (- 1) as (+ 1 - 2):

[tex]\displaystyle k > (k + 1) + {{1}\over{k + 1}} - 2[/tex]. The RS can be written as a square:

[tex]\displaystyle (k + 1) - 2 + {{1}\over{k + 1}}[/tex]
[tex]= \left(\sqrt{k + 1} - {{1}\over{\sqrt{k + 1}}}\right)^2[/tex]. Notice that the terms inside the brackets have a positive value. Therefore,

[tex]\displaystyle \left(\sqrt{k}\right)^2 >\left(\sqrt{k + 1} - {{1}\over{\sqrt{k + 1}}}\right)^2[/tex]. Taking square root of both sides, and moving terms around, we obtain the desired inequality.

nr, we are trying to prove that:

[tex]\sqrt{k}+\frac{1}{\sqrt{k+1}} > \sqrt{k+1}[/tex]

Following what Joe did:

[tex]\frac{(\sqrt{k+1})(\sqrt{k})}{\sqrt{k+1}}+\frac{1}{\sqrt{k+1}} > \sqrt{k+1}[/tex]

Multiply both sides by [tex]\sqrt{k+1}[/tex], assuming that it is true:

[tex](\sqrt{k+1})(\sqrt{k}) + 1 > k+1[/tex]
[tex](\sqrt{k+1})(\sqrt{k})> k[/tex]
[tex]\sqrt{k^2+k}>k[/tex]

Now since [tex]\sqrt{k^2} = k[/tex] for k positive, if you add a k to the inside of the square root:

[tex]\sqrt{k^2+k}>k[/tex]

Guys, is this correct? Danny do you follow all this?

It's correct, but only because each step is reversible (as Joel mentioned). It is best to start with an universally true statement, and manipulate it to arrive at the target inequality, not backwards.

Osiris, I have to admit, what you did is very cool, but I would never think to do any of that...

I did follow what you did!  But Osirsis is, um, wayyy too advance for me.   But here I go stumbling along with matematical induction.

4)

n! > 2^n

Substitute n for 4.

4! > 2^4
24 > 16 <--- Correct.

Assuming that is correct, let's prove the statement for k + 1.

2^n (k+1) > 2^k+1

Divide both sides by 2^n to get:

n+1>2

Substract 1 from both sides.

n > 1 Which is correct since n has to be 4 or greater.

... Right?

5)

Substitute n for 1, x for one, and y for two.

(1/2)^2 < (1/2)^1
1/4<1/2  <--Correct.

Assuming this is correct, let's prove this statement for k+1.

(x/y)^n (x/y)^n+2 < (x/y)^n+1

Multiply together to get:

(x/y)^2n+2 <(x/y^n+1

Divide both sides by x/y^2n+2 which gives you...

1<1/2...Um, where's my stupid mistake?

6)

Let n = 1

2^2(1)-1 + 3^2(1)-1
5

Which is divisable by 5.

Now how do I do this part?

(5 + 2^2n+2 + 3^2n-2)/5?

(4)[quote]Assuming that is correct, let's prove the statement for k + 1.

2^n (k+1) > 2^k+1
[/quote]

You should not have both n and k in the equation.

You assume that [tex]k!>2^k[/tex] when [tex]k \ge 4[/tex].

Now you are going to show that:
[tex](k+1)! > 2^{k+1}[/tex].

I don't see how what you did is correct, cause [tex]2^k[/tex] is not on both sides...or am I mistaken?

I would try:

[tex](k+1)k! > 2^k \cdot 2[/tex]

Now we already assumed that [tex]k! > 2^k[/tex], and you are left w/ [tex]k+1 > 2[/tex], or [tex]k > 1[/tex] as you mentioned, and this is true, so the statement must be true.

I am not great at proving stuff, but I think this is okay (maybe).

I am not exactly sure what you were doing in the beginning steps.  Did you divide the new expression [tex](k+1)! > 2^{k+1}[/tex] by the assumed expression [tex]k! > 2^k[/tex]?  I think you did it pretty well regardless, good work, keep up the effort.

(5) Try this:

Assume that [tex] (\frac{x}{y})^{k+1} < (\frac{x}{y})^{k}[/tex]

What does multiplying both sides by the constant [tex]\frac{x}{y}[/tex] produce?

Your work:

[quote](x/y)^n (x/y)^n+2 < (x/y)^n+1
[/quote]

What you want to show is not this, letting [tex]n=k+1[/tex],

[tex](\frac{x}{y})^{k+2} < (\frac{x}{y})^{k+1}[/tex].


function
topology
real analysis
real analysis unsolved

$f: R^2 \to R^2$ a continuous function such that $f(f(X))=X$ for any $X \in R^2$.
Prove that $f$ has at least on fixed point.
And in $R^n$

Sorry about that. I got confused during that post. I've deleted my message.

What I really meant was this: given a closed path $u$ in $\mathbb R^2$ homotopic to a constant path, and a continuous function $g: \mathbb R^2\to\mathbb R^2$, the closed path $g\circ u$ is homotopic to a constant path in $g(\mathbb R^2)$. In this sense, we have a kind of "preservation of simple connectedness". In the above post I attempted to do precisely this: find a nullhomotopic closed path which is no longer nullhomotopic after applying $g$, thus getting a contradiction. Anyway, since it wasn't really a solution, it's beter off deleted :).

I'll repost it when/if I feel like making everything precise :).

I had read your proof and thought it worked, because $f$ is an homeomorphism (while $exp$ not).
The idea was beautiful  :)

No, cimabue was right. What I said there was indeed wrong, but luckily, as I wrote in my second post, it was not what I meant :). I did not apply those considerations to $f$, but to $g$, which was defined by $g(x)=x-f(x)$, and $g$ is not necessarily a homeomorphism.

Consider a segment $[0,1]$. A point $a\in [0,1]$ is called good, if $f(a)$ belongs to $[0,1]$. The set of good points is closed, so we may find a good $b$, for which the distance between $b$ and $f(b)$ is minimal. The segment $\delta$ between $b$ and $f(b)$ does not intersect with its $f$-image $f(\delta)$. The union of $\delta$ and $f(\delta)$ is a Jordan curve $\gamma$. We have $f(\gamma)=\gamma$. The interior $D$ of $\gamma$ is connected. If there are $x\in D$ and $y\in D$ such that $f(X)\notin D,\,f(Y)\in D$, then by connectness we find $z\in D$ such that $f(z)\in \gamma$ --- a contradiction. Analogously, the $f$-image of the exterior $D'$ of $\gamma$ either lies inside $D$, or inside $\D'$. Since $f$ is bounded on $D$, $f(D')$ contains some points in $D'$ (any point in $D'$ has a preimage). Finally, $f(D)=D,\,f(D')=D'$.

Now apply Brower theorem for $D\cup \gamma$. I do not know exactly, whether $D\cup \gamma$ is homeomorphic to a closed disc, but I remember there was some theorem about this.

has someone an idea in higher dimensions ?


geometry
trigonometry
symmetry
trig identities
Law of Sines
Law of Cosines
angle bisector

Here's a challenge:

Given:

ABCD is a square
Point M is inside ABCD
Angle MDC = Angle MCD = 15 degrees

Prove:
Triangle ABM is equilateral

Hints:
[hide]You need to construct one line[/hide]
[hide]Uses tangent and tangent inverse[/hide]

If any one knows a different way to solve this problem that doesn't use the hints above, then please post, I would like to see it.

Though it's not necessary to use only [hide] tangent [/hide], I think the only solution is trigonometric. :D

yeah.  because you can use

[hide]sin
by eliminating the possiblity of scalene and doing some other stuff, but i don't remember that way.[/hide]

Eh. I found that it was isoscoles.
I assumed it was equilateral so then I found the altitude of the triangle in terms of the square's side and then I just used trig to find the length of the altitude of the 15-15-150 triangle in terms of the side of the square...adding them up the answer should be the length of the square, right...? I found that it was something different...
And it's obviously not scalene.

its equilateral, i don't know what you did...

err ok.
So then the altitude of the triangle is $\frac{s\sqrt{3}}{2}$. So then with the law of sines, $\frac{\sin75}{\frac{s}{2}}=\frac{\sin15}{x}$ where $x$ is the altitude of $\triangle MDC$. Solving for $x$, $x=\frac{s\sin15}{2\sin75}$. Then add: $\frac{s\sin15}{2\sin75}+\frac{s\sqrt{3}}{2}$, take out $s$ and get $s\displaystyle\left(\frac{\sin15}{2\sin75}+\frac{\sqrt{3}}{2}\right)$ and then I used the trigonometric identities (and a calculator afterwards), and if $\triangle ABM$ is equilateral, then $\frac{\sin15}{2\sin75}+\frac{\sqrt{3}}{2}=1$...oops. It does. hehe, ABM is equilateral..

[hide="my trig way"]
ok draw a line through $M$ parallel to $AD$ and $BC$.  So the leg of the isoceles triangle $MCD$ will be $\frac{s}{2\cos{15}}$ by simple trig.  Then we have $\angle ADM = 75$ because square has right angles.  then we have find $\cos{15}$ and $\cos{75}$ by sum-difference formula or other ways you want to, and using law of cosines and simplify heavily we have that $AM = s$, and by symmetry $CM = s$, therefore it'll be an equilateral triangle
[/hide]

Correct me is im wrong:

Given DM=CM, angle ADM=MCB, and AD=BC. Therefore triangle ADM is congruent to BMC. So AM=BM. Now i just need to prove AB equals AM or BM.

Draw Line EM, so that EM is the angle bisector of triangle AME intersecting AB at E. Note that the angle bisector, the median and the altitude of a iscoceles triangle are the same line. Draw line FM so that it is the angle bisector of triangle DMC, intersecting CD at F. Call AD, a so that a=EM+FM. FM=a/2*tan 15. Therefore EM=(1-tan15/2)a, which equals sqr3/2. Notice that AE=EB=DF=FC=a/2. So ((1/2)^2+(sqr3/2)^2)a=AM. Which proves AM=AD=BM.

Sorry i cant get a picture. :oops: [/list]

[quote="star99"]but wouldnt that mean that DN = DM = MN = AN  and since AM = AD, then AN+ND = AD - which it cant be since it wouldnt be a triangle.

there is either something wrong with my logic or yours...[/quote]
How did you get $MN=AN$? :?

Alright, I'll post my solution based on the hint I've posted:
[hide="Geometric solution"]
By ASA, $\triangle AND$ and $\triangle DMC$ are congruent., so $DN=DM$.
Also, $\angle NDM=90^\circ-15^\circ-15^\circ=60^\circ$, which implies $\triangle MDN$ is an equilateral triangle.

$\angle AND=180^\circ-2\cdot 15^\circ=150^\circ$ and $\angle DNM=60^\circ$, so $\angle ANM=360^\circ-150^\circ-90^\circ=150^\circ$
By SAS, $\triangle AND$ and $\triangle ANM$ are congruent.
So we have $AD=AM(=AB)$ and $\angle NAM=15^\circ$.
That gives $\angle MAB=90^\circ-2\cdot 15^\circ=60^\circ$.

Since $AM=AB$ and $\angle MAB=60^\circ$, $\triangle ABM$ is equilateral.
[/hide]
Hope it's clear now.

[quote="frt"][hide="Try this :D "]
Let $N$ be a point inside the square such that $\angle ADN=\angle DAN=15^\circ$.

Step 1: Prove that $\triangle DMN$ is equilateral.
Step 2: Prove that $AD=AM=AB$.
Step 3: Prove that $\angle MAB=60^\circ$.

Then we're done! :lol:
[/hide][/quote]

That's genius—I'm kinda jealous I didn't think of it first. Anyways here's a fully written-out non-trig solution based on that.

[hide]
Let $N$ be a point inside the square such that $\angle ADN=\angle DAN=15^\circ$. By ASA, triangles $ADN$ and $CDM$ are congruent. Thus, $DM = DN$. $\angle MDN$ must equal $60^\circ$, because $\angle ADN + \angle MDN + \angle CDM = 90^\circ$. Since triangle $MDN$ is isosceles and one of the angles is $60^\circ$, it's clearly an equilateral triangle. So that means that $MN = DN = AN$. Also, $\angle ANM = 150^\circ$, because $\angle AND + \angle DNM + \angle ANM = 360\^circ \implies 150^\circ + \angle ANM + 60^\circ = 360^\circ$. So triangles $AND$ and $ANM$ are congruent by $SAS$. Thus $AM = AD = AB$. Since triangles $ADM$ and $BCM$ are congruent by $SAS$, $AM = BM$. So our conclusion is that $AM = BM = AB$ and thus triangle $ABM$ is equilateral.

No trig.
[/hide]

Let $N$ be the point in $ABCD$ such that $\angle BAN=30^{\circ}=\angle CBN$.

Then $\angle NAB=\angle NBA=90^{\circ}-30^{\circ}=60^{\circ}=180^{\circ}-60^{\circ}-60^{\circ}=\angle ANB$, so $\triangle ANB$ is equilateral.

$\triangle DAN=\triangle CBN$ by SAS, so $ND=NC$, and $\angle NDC=\angle NCD$.

Then $\angle ADN=\angle AND=\frac{180^{\circ}-30^{\circ}}{2}=75^{\circ}$.

$\angle NDC=90^{\circ}-75^{\circ}=15^{\circ}=\angle NCD$.

$N$ has the same properties as $M$ and they are uniquely defined so $N\equiv M$ giving: $\triangle AMB$ is equilateral.

wow thats a lot of ways.....

n e one have more geometry challenge problems?


trigonometry
inequalities
inequalities unsolved

Given triangle $ABC$ with $A\le 120^{0}$ find minimum of
$\frac{(\sin A\sin B+\sin B\sin C+\sin C\sin A)^{2}}{\sin A\sin B\sin C}$

[quote="arqady"][quote="nayel"]what is the minimum of $\sin A+\sin B+\sin C$?[/quote]
If $A,$ $B$ and $C$ are measureangles of triangle then
$\inf(\sin A+\sin B+\sin C)=0.$ :wink:[/quote]

thank you. but there's a tiny bit of problem: i lost my paper! :blush:

[quote="arqady"][quote="nayel"]what is the minimum of $\sin A+\sin B+\sin C$?[/quote]
If $A,$ $B$ and $C$ are measureangles of triangle then
$\inf(\sin A+\sin B+\sin C)=0.$ :wink:[/quote]
With $A\le 120^{0}$ that express has minimum when $A=120^{0},B=C=30^{0}$, I haven't still solved that this problem.

[quote="nayel"]ok i rediscovered my theory:

we have

$(\sin A\sin B+\sin B\sin C+\sin C\sin A)^{2}$
$\ge 3\sin A\sin B\sin C(\sin A+\sin B+\sin C)$

so we only need find minimum of $\sin A+\sin B+\sin C$

i think that didn't help... :([/quote]
I think you can't do follow your way because follow me my express gets minimum iff $A=120,B=C=30$ but your inequality gets equality iff $A=B=C=60$ :maybe:

The minimum of the expression doesn't get minimum only if $A=120$ because the expression is cyclic so we can chose also $B=120$ or $C=120$.For example we can have $A=20,B=20,C=140$ so we have an angle bigger than $120$ and because the expression is cyclic we can re-denote the triangle so it has no importance that $A\leq120$.We have no restriction.I hope you understood me.

I haven't understood you Zamfirmihai, my express won't be symmetric woth condition $A\le 120$, by the way really I sure my problem is true minimum when $A=120,B=C=30$, it is from a problem follow me is very hard
Given triangle $ABC,A\le 120^{0}$ prove that
\[(FA+FB+FC)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge 4+\frac{4}{\sqrt{3}}\]
With $F$ is Fermat point Equality holds iff $A=120,B=C=30$ two problems is the same. but I am sure the second is true, symmetric isn't problem here.


How man seconds will it take a train 300 meters long travelling at 100 km/hr to pass a man jogging at 10 km/hr in the same direction?

[hide]You want the front of the train to start at where the man is, and then travel 0.3 kilometers.

x = the number of hours

10x+0.3 = 100x

0.3 = 90x

x = 0.00333...

But x is the number of hours.  We want the number of seconds.

0.0333.../x = 1/3600

x = 12

So it would take 12 seconds for the train to pass.

[/hide]

[hide]The train is going 100-10 = 90km/h relative to the man so,

 0.3km/ 90km/h= 1/300h = 3600/300s = 12 seconds[/hide]


summer program
MathPath
ceiling function
AMC
USA(J)MO
USAMO
email

Who are the freshmen who made it to MOSP?  Try to include name and state, and score maybe to help those in waiting.

I'm Jason Hoch, CT, 9

uh...I think they'll send out a complete list pretty shortly

John Doe, AK, $9 \frac{3}{4}$. I think they will be sending out a list shortly too, but I guess people are just really excited and want to share :P.

[quote="probability1.01"]John Doe, AK, $9 \frac{3}{4}$.[/quote]

I assume that rounds up, right? :D

[quote="nr1337"][quote="pieterminate"]41!!!???  Some rounding!  :D  Yeah, so congrats to making red mop with $9\frac 34 =41$  :lol:[/quote]

Well actually it's $\left\lceil 9\frac 34 \right\rceil = 41$.  So $10 = 41$, which is true if the LHS is in base $41$ and the RHS is in base $10$.

No more math until MOSP! :P[/quote]

10 is *congruent* to 41 mod 31!  :lol:

[quote="MysticTerminator"]you have to complete every problem (approximately USAMO level difficulty; some towards the end are a bit harder) or you'll be cut from the program[/quote]
and, I might add, you get sent home on a Metro Bus.... or ride a bike, you're pick. They will provide you with a week's worth of sacked lunches if you choose the latter option, at no expense to your family (but they will charge you for the bike :D)

Oh, and I completely forgot that probability1.01 counts in base $\frac{7+\sqrt{689}}{64}$. Then his USAMO score is 9.75   :?

[quote="pieterminate"][quote="MysticTerminator"]It's actually MOP, and you have homework - you have to complete every problem (approximately USAMO level difficulty; some towards the end are a bit harder) or you'll be cut from the program.[/quote]

Eh, [b]m[/b]ath [b]o[/b]lympiad [b]s[/b]ummer [b]c[/b]amp

But are they actually that hard....?
And has [i]anyone[/i] ever been cut?  :roll:  :?  ;)[/quote]

um, that would be MOSC, which is obviously wrong - it's actually Math Olympiad Program, hence the name MOP and the corresponding MOPpers (not MOPsters, MOSPers, anything else stupid)

yes, if you use MOSP (ugh, feel dirty even typing it), you're automatically an outcast and get sent home forcibly with all the people who didn't do their homework

[quote="MysticTerminator"][quote="nr1337"][quote="MysticTerminator"]you have to complete every problem[/quote]

Note how he cleverly avoids the word "correctly..." :)[/quote]

?! well of course - if you do ANYTHING at MOP incorrectly, ... (you don't just get sent home)[/quote]

Something tells me that you also get some unimaginable punishment by the [b]MOP[/b] people if you qualify but don't go, right... :D

[quote="pieterminate"]
But are they actually that hard....?
And has [i]anyone[/i] ever been cut?  :roll:  :?  ;)[/quote]

Ummm... obviously? But of course if you don't believe me you can always go there and find out for yourself ;). So yeah, don't get too excited yet.

Oh well, I was just reading from the email:
[quote]
The Mathematical Olympiad Summer Program
(MOSP)[/quote]

Yeah, you people are scaring me too much. :roll:  :lol:  lol
Oh well, that makes MOP more fun then.  :D  :lol:

[quote="pieterminate"]Oh well, I was just reading from the email:
[quote]
The Mathematical Olympiad Summer Program
(MOSP)[/quote]

Yeah, you people are scaring me too much. :roll:  :D  :lol:  lol[/quote]

that's b/c the AMC has changed the name from MOP to MOSP, but only losers actually call it MOSP.  In fact, if you are going to use the "S" at all, you can't capitalize it or else you will be crucified.  So it's MOP or MOsP, but preferably just plain old MOP.

Don't worry; when they cut you from the program, they don't actually send you home.  Instead, MOP just gives you a mop, and you work there as a janitor ;)

In all seriousness, if you don't want to go to MOP with a mop, you better complete EVERY homework problem.  Otherwise, you might end up like what almost happened to Alex last year :(
(sorry, alex, I know you didn't want me to mention it... I didn't give any details though  :oops: )

[quote="bubala"][quote="pieterminate"]Oh well, I was just reading from the email:
[quote]
The Mathematical Olympiad Summer Program
(MOSP)[/quote]

Yeah, you people are scaring me too much. :roll:  :D  :lol:  lol[/quote]

or else you will be crucified.

Don't worry; when they cut you from the program, they don't actually send you home.  Instead, MOP just gives you a mop, and you work there as a janitor ;)
[/quote]

How cruel! :rotfl:

How hard are the problems exactly?  Stupid people like me might not be able to solve them... :D

[quote="nr1337"]I heard that the mops are even colour-coded: Freshman get red mops, winners get black mops, and everyone else gets blue mops. :D[/quote]

wait . . . how did you know? (actually, on the '04 name tags, the shirt of the guy with the mop was indeed colored that way)

Word is Edward Trout made it too. And a couple of Japanese people but I don't remember how to spell their names.

[quote]How hard are the problems exactly? Stupid people like me might not be able to solve them...[/quote]
Here's a sample problem:

[quote]Prove or disprove Goldbach's conjecture.[/quote]

Oh if they're at that level i'll be all set i started toying with that a couple years ago

[quote="nr1337"][quote="EFuzzy"]Edward Gan, MD, 12[/quote]

Hey, I saw you at MathPath! :D  Congrats![/quote]

Congratulations to you too! Do you know anyone else from Mathpath who might come?

Look for a long-haired Georgian wearing a blue blazer and green vest.

is it possible that students outside the states get invited to the mosp?

[quote="jrshoch"]Anyone know how to change the topic title.... it has the dreaded letter S in it.[/quote]

That's easy. The person who posted the first post in the thread just has to change the subject line of his post with an edit.


Let x,y,z be positive numbers such that $ x^2\plus{}y^2\plus{}z^2\equal{}6$
and A,B,C are three angles of an acute triangle.
Prove that
 $ \sum_{cyc}\frac{1}{1\plus{}yzcosA\plus{}xyz^2cosAcosB} \geq 1$

[quote="8826"]Let x,y,z be positive numbers such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 6$
and A,B,C are three angles of an acute triangle.
Prove that
 $ \sum_{cyc}\frac {1}{1 \plus{} yzcosA \plus{} xyz^2cosAcosB} \geq 1$[/quote]
Use: $ \sqrt{3\sum \ {xyz^2cosAcosB}} \le \sum \ yzcosA \le \frac{\sum \ x^2}{2}$ and Cauchy-Schwarz inequality,we have Q.E.D


function
geometry
3D geometry
calculus
calculus computations

$ f(A,B,C,D)$ is continuous and convex function, $ A,B,C,D \in$ in a cube.
Is it true that we have the maximum of $ f(A,B,C,D)$ when $ A,B,C,D$ lie in the vertices of the cube?

A hypercube, but yes.  To see this, consider each of $ A, B, C, D$ separately.

No.
The problem is this.
Consider a cube on $ R^3$, and $ A,B,C,D \in$ cube.
Find the maximum of $ f(A,B,C,D) \equal{} AB \plus{} BC \plus{} CD \plus{} AD$.

[quote="Goxer"]Consider a cube on $ R^3$, and $ A,B,C,D \in$ cube.[/quote]

I'm not sure what you mean.  I originally interpreted $ (A, B, C, D) \in \mathbb{R}^4$.  Do you mean that $ A, B, C, D$ are points in $ \mathbb{R}^3$?  Does $ AB$ denote the dot product?

$ A,B,C,D$ are point in $ R^3$.
$ AB$ is the euclidean distance.

For your example: the maximum occurs when $ A\equal{}C$ and $ B\equal{}D$, and these pairs are opposite corners of the cube.

Hmm.  Well, I'll tell you what I originally meant.

[b]Proposition:[/b]  Let $ f : \mathbb{R}^n \to \mathbb{R}$ be a continuous and convex function defined on a hypercube (or a cube, or a square, or an interval).  Then $ f(\vec{v})$ attains its maximum when $ \vec{v}$ is some vertex of the hypercube (etc.).

This problem can be interpreted as a function $ f : \mathbb{R}^{12} \to \mathbb{R}$.  The vertices of the $ 12$-dimensional hypercube correspond to each of $ A, B, C, D$ being vertices of a cube in $ \mathbb{R}^3$, so the original conclusion should still hold as long as $ f$ can be shown to be convex in all $ 12$ variables.

Then the answer initially given by [b]t0rajir0u[/b] still applies. With $ A,B,C$ fixed, you have a convex function of $ D$, which attains its maximum at an extreme point of the cube, that is, a vertex. And so on.


geometry
3D geometry
probability
number theory
least common multiple
function

I'm sorry to tell you that I didn't have time to grade or post any solutions yet.

You guys can discuss it though. I think Drunken Math and 236factorial were only ones who sent the answers but it'll be good discussion for you guys to have.  :)

I'll post the results and solutions ASAP.

It looks like i lost the original copy, so you'll have to read a PDF  :D

[hide="warning"]some are wrong   :D (like 10 i think) [/hide]

Results should be posted soon but I can't give anyone exact date. I have practices everyday and since I'm sophomore, I have more school work to do than last year (and I'm taking higher classes than my age) so be patient (I know everybody is and I appreciate this greatly) and I'll grade them ASAP.

Thanks.

Here are my answers (from the PM I sent Silverfalcon):

[hide]1.) Figuring out the first few terms of the sequence,

3, 5, 9, 17, 33, 65, 129...

There's a pattern on the units digits of the terms of this sequence; it goes 3, 5, 9, 7 repeatedly.

Since 20 is divisible by 4, the units digit of 20 must be $\boxed{7}$.

2.) $h(4)=(2*4)(4+3)=8*7=\boxed{56}$

3.) To maximize the sum, we must make abc have the largest possible value that satisfies the conditions. Since there are no conditions restricting the value of b, we make the largest we can make it: 9. a and c can't be equal to each other and must be smaller than c, so the largest possible values of those are 8 and 7. It doesn't matter what order they are in, since the brother number is the number flipped. So the largest sum of $abc$ and $cba$ is $897+798=\boxed{1695}$

4.) Set T has these numbers: 66, 55, 44, 33, 22, and 11, all in base 7. Converting them into base 10 would yield Set R. Set R has these numbers: 8, 16, 24, 32, 40 and 48. Out of those 6 numbers, the only two that are a perfect square or a perfect cube are 8, and 16. So the probability would be $\frac{2}{6}=\boxed{\frac{1}{3}}$

5.) Counting the number of possibilities, the number of possibilitites is:

$(6+5+4+3+2+1)+(5+4+3+2+1)+(4+3+2+1)+(3+2+1)+(2+1)+1=\boxed{56}$

6.) $n+n+1+n+2=3n+3=3(n+1)$ The LCM of 3 and 4 is 12, so we must make $n+1=4$. So $\boxed{n=3}$

7.) If you try out different digits for the thousands digit, you will find that there are $\boxed{6}$ different numbers that satisfy the condition.

8.) Do notice that if you divide $12^1$ by 13, the remainder is 12. If you divide $12^2$ by 13, the remainder is 1. If you proceed with casework, you find that this is a neverending pattern, in which if the exponent is odd, the remainder would be 12, and if the exponent is even, the remainder is 1. Since 12 is odd, the remainder would be $\boxed{1}$

9.)

10.) The largest circle that can be inscribed in a unit square has a radius of $\frac{1}{2}$, so the area of that circle would be $\boxed{\frac{\pi}{4}}$[/hide]

[quote="Drunken_Math"]10.) The largest circle that can be inscribed in a unit square has a radius of $\frac{1}{2}$, so the area of that circle would be $\boxed{\frac{\pi}{4}}$[/quote]

I knew i should've put that... I got tricked  :D or did I?  :?

Since you guys seem to be anxious about your score, I'll post the scores first. Both of you qualified!!

Drunken_Math: 9 correct
263factorial: 8 correct

Since the solution is pretty long and important, I'll post it on the next post.

Here is a solution for this COPSD 3rd A.

[hide="Solution 1"]
There are two methods for this one.

Method 1:

$t_1 = 3$ and $t_2 = 5$. Therefore:

$t_3 =  3(5) - 2(3) = 9\\
t_4 = 3(9) - 2(5) = 17\\
t_5 = 3(17) - 2(9) = 33\\
t_6 = 3(33) - 2(17) = 65\\
t_7 = 3(65) - 2(33) = 129$

Thus, the last digit repeats in every 4 terms. The last digit of $t_{20}$ is then 7.

Method 2:

[b]Lemma: For $n>2$ in $t_n$, $t_n = 2^n + 1$ [/b]

Proof: For $n = 3$ (Let's start there although $n = 1,2$ work), we get $t_3 = 2^3+1 = 9$ which is the same number from Method 1. Now, assume for some integer $k$, we have $t_k = 2^k + 1$. Hence, $t_{k+1}$ will be:

\[{t_{k+1} = 3(2^k + 1) - 2(2^{k-1} + 1)\\
= 3+3(2^k} - 2^k - 2\\
= 2^{k+1} + 1\]

which implies $t_k = 2^k + 1$ works for all natural $k$ values. Thus, $t_{20} = 2^{20} + 1 = 1048577$ and 7 is indeed the last digit. [/hide]

[hide="Solution 2"]

First, put 4 into each function:

$f(4) = 2 \cdot 4 = 8\\
g(4) = 4+3 = 7$

$h(4)$ is simply product of those two which is 56. [/hide]

[hide="Solution 3"]

To maximize $abc+cba$, we note that this equals:

$10^2(a+c) + 10^1(2b) + 10^0(a+c)$

Thus, $a+c$ has to be maximum in order for whole thing to be maximized. Since $a \neq c$ and $b > a,c,$ we can have $a = 8$ or $c = 8$ for maximum (which means $b = 9$!). So, $10^2(a+c) + 10^1(2b) + 10^0(a+c) = 15(100) + 18(10) + 15 = 1695$. [/hide]

[hide="Solution 4"]

Let's simply say numbers that have identical unit and tens digit as palindromes. Six palindromes in base 7, or elements of set $T$, are 11,22,33,44,55,66. Now, when we convert them to base 10:

$11_7 = (1 \cdot 7^1 + 1 \cdot 7^0)_{10} = 8_{10}\\
22_7 = (2 \cdot 7^1 + 2 \cdot 7^0)_{10} = 16_{10}\\
33_7 = (3 \cdot 7^1 + 3 \cdot 7^0)_{10} = 24_{10}\\
44_7 = (4 \cdot 7^1 + 4 \cdot 7^0)_{10} = 32_{10}\\
55_7 = (5 \cdot 7^1 + 5 \cdot 7^0)_{10} = 40_{10}\\
66_7 = (6 \cdot 7^1 + 6 \cdot 7^0)_{10} = 48_{10}$

Only 2 are square or cube so the probability is $\frac{2}{6} = \frac{1}{3}$. [/hide]

[hide="Solution 5"]

This problem is same as finding number of ways put $( \star )$ in $( * )$ spaces where $( \blacksquare )$ are apples arranged in a way such that no two bananas (or $\star$ ) can be next to each other.

$* \blacksquare * \blacksquare * \blacksquare * \blacksquare * \blacksquare * \blacksquare * \blacksquare *$

There are 8 $*$'s and 3 $\star$ we need. There are $\binom{8}{3} = 56$.  [/hide]

[hide="Solution 6"]

$n+n+1+n+2 = 3n+3$. In base 4, this equation is $3n + 3 \equiv 0 \equiv 4 \equiv 8 \cdots \mod 4$. Ths smallest $n$ is if $3n+3$ or $3(n+1)$ is divisible by 4, and obviously, at $n = 3$. [/hide]

[hide="Solution 7"]

To be divisible by 15, it must be divisible by 5 AND 3 as well. So, we have two cases:

Case I: A790

Since $3|A+7+9$, $A$ can be 2,5,8.

Case II: A795

Since $3|A+7+9+5$, $A$ can be 3 (not 0 because $A$ is the first digit!), 6, 9.

Hence, there are 6 $A$ values or such numbers. [/hide]

[hide="Solution 8"]

$12^{12} \equiv (-1)^{12} \mod 13\\
\equiv 1^{12} \mod 13\\
\equiv 1 \mod 13$

Thus, the remainder is 1. You can also solve this by seeing that remainder alternates between 1 and 12. [/hide]

[hide="Solution 9"]

$pq = \frac{p}{q}\\
q = \frac{1}{q}\\
q^2 = 1\\
q = \pm 1$

For $q = 1$, we get $p-1 = p$ and that implies $-1 = 0$ which is false. For $q = -1$, we get $p+1 = -p$ which implies $p = \frac{-1}{2}$. So, $k = -1 \cdot \frac{-1}{2} = \frac{1}{2}$. [/hide]

[hide="Solution 10"]

The largest square is one that contains the 4 midpoints of four sides of square (draw it and you'll see it!). Thus, the inscribed circle has radius of $\frac{1}{2}$, and it will have area of $\frac{\pi}{4}$. [/hide]

[quote="236factorial"]Number 7.... forgot could end with 0... so dumb... I suck at math  :D[/quote]

Don't worry, everybody makes their share of careless mistakes.  :)

And you... skipped number 9?!  :o

It's better to try and get it wrong than to not try at all.. Santa thinks you're naughty  :rotfl:

Of course, that means everyone is naughty  :D

$p-q = pq =\frac{p}{q} = k$

Here was my reasoning for this problem:

$pq=\frac{p}{q}$

$pq^2=p$

$q^2=1$

$q=1$

I forgot to make it $\pm1$

As I said, everyone makes their share of careless mistakes.  :)

I'm glad to see some kind of talk is going on.

I wasn't sure if all my hard work typing the solutions attracted nobody.  :P


geometry
3D geometry
dodecahedron
pigeonhole principle

prove that choosing any 5 vertexes of a regular dodecahedron three of them will form an isosceles triangle

[hide]From any 5 points, there are $ \binom52\equal{}10$ lengths.  However, there are only 6 different lengths for diagonals possible, so thus, by the pigeonhole principle, at least two of them must be the same length and thus, an isosceles triangle can be formed by choosing the vertices used to create those two sides.[/hide]

[quote="b-flat"][hide]From any 5 points, there are $ \binom52 \equal{} 10$ lengths.  However, there are only 6 different lengths for diagonals possible, so thus, by the pigeonhole principle, at least two of them must be the same length and thus, an isosceles triangle can be formed by choosing the vertices used to create those two sides.[/hide][/quote]

actually there are 7 different lengths possible, because the sides of the triangle don't have to be diagonals. but the results is the same anyway


Solve the equation:
$ \sqrt{x \minus{} 3} \plus{} \sqrt{7 \minus{} x} \equal{} \minus{}x^{2} \plus{} 6x \minus{} 7$

here is my solution:
$ \minus{}x^2\plus{}6x\minus{}7<\equal{}2$ and $ \sqrt {x \minus{} 3} \plus{} \sqrt {7 \minus{} x} > \equal{} \sqrt{4} \equal{} 2$ so x=3 is the solution

[quote="Nam Luu"]Solve the equation:
$ \sqrt {x \minus{} 3} \plus{} \sqrt {7 \minus{} x} \equal{} \minus{} x^{2} \plus{} 6x \minus{} 7$[/quote]

Put $ y\equal{}x\minus{}3$ the equation $ \Leftrightarrow \sqrt{y}\plus{}\sqrt{4\minus{}y}\equal{}2\minus{}y^2$

Now $ LHS^2 \ge 4 \Rightarrow LHS \ge 2 \ge RHS$

Equality holds $ \Rightarrow y\equal{}0 or 4$ (rejected)

So $ y\equal{}0 \Rightarrow x\equal{}3$ is the only solution after checking.


Let $XOY$ be an angle formed by lines $OX,OY$.Let$P$ be a point, inside the angle.
Find the line $\ell$ such that it passes through $P$ and meets $OX,OY$ in $A,B$ such that $AP*PB$  be minimum.

I believe it's the line passing through $P$ which makes equal angles with $OX,OY$. In other words, $OAB$ should be isosceles.

Take $AB$ as above (in order to make $OAB$ isosceles), and let $A'B',\ A'\in OX,B'\in OY$ be some other line through $P$. We want to show that $PA'\cdot PB'\ge PA\cdot PB$.

Take $A''\in OX,B''\in OY$ so that $\angle A''PA=\angle A'PA=\angle B''PB=\angle B'PB$ ($A'',A'$ are separated by $A$, and $B'',B'$ are separated by $B$). It's easy to see that the quadrilateral $A''A'B''B'$ is cyclic, and let its circumcircle cut $(PA,(PB$ in $A_1,B_1$ respectively. Then $PA_1\ge PA,PB_1\ge PB\Rightarrow PA'\cdot PB'=PA_1\cdot PB_1\ge PA\cdot PB$, as required.

Exactly grobber,here is a another solution. which i really like: :D
Let $ABO$ be an isosceles trinagle such that $P \in AB$.
Let $C$ be a circle ,inside the $XOY$ and tangent to $OX,OY$ at $A,B$.
We claim that $PA*PB$ is the minimum.(note that It is power of $P$ wrt $C$).
Draw another line which intersects $OX,OY$ in $R,S$.
Clearly we have $RP*SP \ge$ power of $P$ wrt $C$.
Problem solved.

PS.Its a reall pitty that you guys dont have goemetry in university. :(


number theory
greatest common divisor
probability
relatively prime
number theory proposed

What is the probability that a random 2 digit base b number will be relativly prime with its digit reversal?

By direct calculation there are 41 two-digit integers relatively prime to their reversal, counting 10.  Thus, if you wish to include integers ending in 0 (whose reversal has only one digit), the answer is $ \frac{41}{90}$.  If you want to exclude these numbers, the answer is $ \frac{40}{81}$.

The numbers which are not relatively prime to their reversals can be characterized as:
1) Numbers with both digits even
2) Numbers in which the first and last digit are both 5 or 0
3) Palindromes
4) Multiples of 11
5) Multiples of 3
6) 70

There is substantial overlap among these categories for 2-digit numbers (e.g. (3) and (4) are actually the same), but for higher-digit numbers each group will be distinct and this characterization may no longer be comprehensive.

Edit: in the 3-digit case, excluding numbers which end in 0, the first five sets form a complete characterization with the exception of the numbers 259 and 952 which have a GCD of 7.  In higher-digit cases, the number of neq categories needed will increase.  In fact, each prime will have some list associated to it of pairs of integers which are reversals of each other and divisible by that prime, and it is most likely not straightforward to characterize these lists as it is for 2, 3, 5 and 11.

Thank you for the reply. I wrote a python script to find the probability for higher numbers of digits(Up to 10000), and it seems to hover around .43 . That is odd, as your characterization of numbers that are relatively prime with there reversal almost suggests something more intricate. I guess I'll have to study the problem further.


probability
function
integration
real analysis
real analysis unsolved

Consider the set $F$ of functions defined as follows $f(0)=0,f(x)=f(i)+\alpha_i(x-i)$ for $i \leq x \leq i+1, 0 \leq i \leq n-1 $ and $\alpha_i$ takes values $-1$ or $1$. Find the probability that for arbitrary taken function $f$ from the set $F$
a).$\int_0^n f(x)dx=0$.
b). the function $f$ has $k$ roots on $(0,n]$

[quote="pleurestique"]a) and b) simultaneously?[/quote]
not, a) is one problem and b) is another.


There is a strange island where people who can climb trees always tell
the truth, and people who couldn't be bothered always lie. You go up
to four people on that island and ask the first person, "Can you climb
trees?" The first person nodded. "Yes," he said. You go to the second
person and ask him. He answers, "I can't climb trees." You go to the
third person, who says, "As a fact, I can tell you that the first and
second person are both liars. I can climb trees." The fourth person
tells you, "I can climb trees, and I can tell you that the third
person is a liar."

One of these people gave an impossible answer, given the information
in the first sentence. Who was it?

Apart from the person who gave you the impossible answer, tell which
person is a truth-teller and which one is a liar.

I'm not sure this is right, but the third person gave the impossible answer because in that island you can't lie.  Uh...I THINK the third person is a lier and the 4th is the truth-teller.

I <b> know </b> I'm wrong.  (I think, who knows!)

That is incorrect. People who [b]don't[/b] climb trees always lie. I've leave this up for a week or so, and if no one else replies, I'll reveal the answer.

The second person gave an impossible reply. If he was a liar, he would say he could climb a tree and if he was a truth teller, he would still say he could climb a tree.

Third person is a liar and the first and fourth person are truth tellers.

[quote="isabella2296"]Here is another problem. It's similar, but also tricky.

A princess visits an island inhabited by two tribes. Members of one tribe always tell the truth, and members of the other tribe always lie.

The princess comes to a fork in the road. She needs to know which road leads to the castle so as to avoid the fire-breathing dragon and rescue the prince from the wizard holding him captive in the castle. (Although the princess doesn't know it, the south road leads to the castle and the north road leads to the dragon.)

Standing at this fork in the road is a member of each tribe, but the princess can't tell which tribe each belongs to. What question should she ask to find the road to the castle?[/quote]

"Show me where you come from." :D

Well...I guess that could work but it's not really a question...

By the way, there are a couple answers that could work. I'll leave this up for a while and if no one else offers any answers I'll give one.

It's a classic problem. The princess should ask [hide]"If I ask the other islander which is the right road to the castle, what will he answer?[/hide]
I don't know any other possible solution, so I'm interested in the second answer.

Goodbye
GM

[quote="isabella2296"]Since the question doesn't limit you to one question, you could ask something like "Is the sky blue?" and then ask which way to go.[/quote]

Of course, the sky is not blue, it just appears blue :lol:

Then let me rephrase that: "Does the sky appear blue?"

Better?  :D

Usually the puzzle says "one question only", and in that case the "is the sky blue" scheme doesn't work.

I guess. Well, I'll post another logic problem.

There are three people in front of you. You know that:

One of them is God. He knows everything, and always tells the truth.
One of them is the Devil. He also knows everything, but lies.
The third person knows nothing, but answers questions as if he knows
the answers. His answers, however, are completely useless and could be
right or wrong.

You can ask a total of three questions that can correctly be answered
with yes or no, each to one of the persons. You may choose whom to ask.

Determine who is who...

I'm not sure...

[hide="answer"]Ask: "If i asked you yesterday if (a certain one of the three) was the third person, would you have answered yes?"

The Devil will be forced to tell the truth because he would have said a lie yesterday; he has to lie about that.

If at least two of them say yes, ask the trademork "Does the sky appear blue?" and we shall get who is who now that we have narrowed it down to God and the Devil.

If at least two of them say no (With the first question, God and the Devil must agree), we repeat the question on one of the other people..."If i asked you yesterday if (a another one of the three) was the third person, would you have answered yes?"
What the majority answers will be correct. Then we go back to the "obvious" question stage[/hide]

Yes, that's correct. Congrats, I didn't get that one on my own.  :10:

the third person gave an impossible answer...he is a liar and the fourth person is telling the truth


$ a\equal{}p^k\minus{}b^2$ and $ p^k|(p^k\minus{}b^2)^2\plus{}b\to p^k|b(b^3\plus{}1)$. If $ p|b$, then $ p\not|b^3\plus{}1\to p^k|b$ contradition with $ p^k>b^2$.
Therefore $ p^k|b^3\plus{}1\equal{}(b\plus{}1)(b^2\minus{}b\plus{}1)$. Obviosly $ (b\plus{}1,b^2\minus{}b\plus{}1)\equal{}(b\plus{}1,3)$. It give $ p\equal{}3,b^2\minus{}b\plus{}1\equal{}3^{k\minus{}1}.$
Only $ b\equal{}2,k\equal{}2,p\equal{}3,a\equal{}5$ is solution.

[quote="Mashimaru"]1. Find every pair of distinct positive integers $ a,b$ such that $ b^2 \plus{} a | a^2 \plus{} b$ and there exists prime $ p$ and power $ k \geq 1$ such that $ b^2 \plus{} a \equal{} p^k$.

2. Does there exist a pair of distinct positive integers $ a,b$ such that $ (b^2 \plus{} a \minus{} 1) | (a^2 \plus{} b \minus{} 1)$ there exists prime $ p$ and power $ k \geq 1$ such that $ b^2 \plus{} a \equal{} p^k$?[/quote]

This is problem 3 of St. Petersburg City Mathematical Oympiad
Looking for problem above, you can find solutions. :roll:

[quote="Rust"]...
Therefore $ p^k|b^3 \plus{} 1 \equal{} (b \plus{} 1)(b^2 \minus{} b \plus{} 1)$. ...$ b^2 \minus{} b \plus{} 1 \equal{} 3^{k \minus{} 1}.$
...[/quote]

Could you please explain this step? :(
I get p=3. then $ 3^{k \minus{} 1}|b^2 \minus{} b \plus{} 1$, then..??

use $ b^2<3^k$ and $ 3^{k\minus{}1}| b^2\minus{}b\plus{}1\equal{}1\mod 2\to b^2\minus{}b\plus{}1\equal{}3^{k\minus{}1}$.


By the expansion of a natural n, in basis 8 , find the criteria of divisibility for 7

A number in base 8 can be written $d_0+8d_1+8^2d_2+\cdots +8^nd_n$, where $n$ is some nonnegative integer and $0\leq d_k\leq 7$ for each $k$.

Since $8^k\equiv 1\pmod{7}$ for all $k$, this number is congruent to $d_0+d_1+d_2+\cdots +d_n$. So the number is divisible by 7 if and only if the sum of its base 8 digits is.

In general, a number is divisible by $n$ iff the sum of its digits in base $n + 1$ is divisible by $n$.

And as a sidenote, you can also use the fact that $n \equiv -1 \pmod{n+1}$ to find divisibility by n+1. For example, is 121 in base 10 divisible by 11? $121 \equiv 1 \times 10^2 + 2 \times 10 + 1 \equiv 1 \times (-1)^2 + 2 \times (-1) + 1$
$\equiv 1-2+1 \equiv 0 \pmod{11}$

[quote="quantum leap"]And on the divisibility rules lists I learned in elementary/middle school, it always said there was no general method like the other numbers to ten...[/quote]

What do you mean?  In base $10$, there are the following methods.

- Methods that require the last $k$ digits for some $k$:  All numbers of the form $2^a 5^b$, with $a, b \in \mathbb{N} \cup \{ 0 \}$.  Most popularly, $2, 4, 5, 8, 10$.

- Methods that involve an easy summation of all digits:  $3, 9, 11$.

- Methods that involve a difficult summation of all digits:  $\{ n | (n, 10) = 1 \}$ (it's just a matter of how hard you want to make your job ;) )

Demonstration:  Everyone says that there is no divisibility rule for $7$.  In fact, there is, but it is annoying to use:

We will denote by $P_p = <a_0, a_1, a_2, ... >$ the pattern of numbers by which you multiply the digits of some number to figure out if it is divisible by $p$ (starting from the ones digit).  For example,

$P_3 = P_9 = <1, 1, 1, ... >$
$P_{11} = <1, -1, 1, -1, ... >$

Now,

$P_7 = <1, 3, 2, 1, 3, 2, ... >$

It is not easy to use, but it works.  In fact, it is trivial to prove

$P_p = <1, b \bmod p, b^2 \bmod p, b^3 \bmod p, ... >$

In some base $b$.  This sequence terminates to zero iff $\exists k : p | b^k$, repeats the $1$ term iff $(b, p) = 1$, and repeats without a $1$ term otherwise.


The cutoff this year was a 27, with a max score of 37, so the scores were lower than last year's, although not by much. Personally, I felt that the exam was significantly harder this year, since anatomy is probably my weakest subject. Although I made the cut, I failed to even break 30 -- my score dropped by 9 points compared to last year.

Good luck to all who will be taking the semifinal exam!

i think u meant "r u given exam papers" :D

[quote="unimpossible"]i feel so guilty.

i qualified with a 27.[/quote]

:wallbash  missed it with 26

Made it with a 35! Without doing eco or biodiversity...time to get to work.

what was the maximum marks??
i mean how much mark exam it was?? :)

It was out of 50, the highest score in the nation being 37.

Thanks..

and could u guys post some  questions ,in free time?? :)

ugh sure? took it today it was INHUMAN ZOMG. fail.

I thought it was okay. I'm hoping for 120 points.

Does anyone else think this exam is flawed? Honestly, the highest score was 37 out of 50. The person with the most knowledge of biology missed 13 questions on a multiple choice test.

I got a 22, literally not knowing any question, and using a little bit of reasoning.

Imho, it could be designed better, but that's just the nature of the beast. The fact that the best score was 37 just means that the test is hard, not necessarily that it is flawed. Differentiation might be somewhat of an issue, but more difficult questions simulate the multiple-choice questions from semis more closely.


quadratics
algorithm
number theory proposed
number theory

Do there exist a method to solve the following one equation without using theory of Pell's equation?

Solve in positive integers : $ 3q^2\minus{}p^2\equal{}2$ if $ p$and$ q$ are both odd.

Define "without the theory of Pell's equation."  One way to derive the Pell's method is to use continued fractions, which is [i]technically[/i] an answer to your question.  In particular, one computes the value of $ 3q_n^2 \minus{} p_n^2$ for convergents $ \frac{p_n}{q_n}$ of $ \sqrt{3}$ and checks if they are equal to $ 2$.  It is known that this sequence is always finite for any quadratic surd, so this algorithm terminates.  Every solution appears in this way.

Could You explain in more details about what is Your method about? I would be really glad if anyone could show me how to kill this equation. Or maybe You know where is an article of method described above? thanks for interests

Briefly:  the convergents $ \frac{p_n}{q_n}$ have the property that $ \left| \sqrt{3} \minus{} \frac{p_n}{q_n} \right| < \frac{1}{q_n^2}$, which can be used to show that $ \left| 3q_n^2 \minus{} p_n^2 \right| < 2 \sqrt{3}$ (and this generalizes).  Other lemmas on continued fractions can be used to show that this sequence is in fact periodic, and what's more has the property that every solution to $ 3q_n^2 \minus{} p_n^2 \equal{} k$ appears for any integer $ |k| < 2 \sqrt{3}$ (this is because all such numbers form a quotient close enough to $ \sqrt{3}$ that it must be a convergent).  This is not true when $ |k| > 2 \sqrt{3}$ - the question is much harder in this regime, but ZetaX knows more than I do here.


:)
Name: V V V  Satyanarayana(Vcube)
residence: Hyderabad
 hello sir
           I want to intract with in maths

:)
Name:George W Bush(dubya)
residence: Texas
hello sir
I want to intract with in iraqi oil

[hide]use the welcome to india thread for intro :spam: [/hide]

Fikar not V^3
They had term everything as spam :D
Welcome!
 :welcomeani: Which class?

[quote="Scary math"]They had term everything as spam :D [/quote]
u forgot 2 mention
SPAMMING IS ALLOWED HERE, EVERYONE SPAMS, EVEN THE MOD  :rotfl:

@Sagar
Abt spamming
there are sections of AoPS where spamming is strictly forbidden,like the FTW forum.Over there,even a wrong solution is considered a spam!!!! :(
@V^3
This is a wried and wonderful place for lost souls.Enjoy!!!!! :thumbup:

@scary
i was talking of the india section only  :)

King of spammers is here!!! :D
Name:James Bond <assistant>
residence:varies

Found :Top secret info on ur computer.Run!!coz i m pursuing u

@ritu
Bond's a spammer???????????? :huh:

I think you guys scared V3 away.... :rotfl:

My intro
Name-Rob Bourdon
Residence-Agoura hills
I wish to interact in rock++

[quote="franklampard8"]course he is!
if he can get a royal flush in every hand in casino royale , y the heck cant he be a spammer?  :P
Anyway here goes
Name: Ajmal Amir Kasab (A2K)
residence: Arthur road Jail
                                  HELLO SIR
[img]http://pics.shareordie.in/2008/12/Mohammed_Ajmal_Amir_Kasab.jpg[/img]
I want to intract with in terrorism :ninja:   :idea:[/quote]
!!   :roftl:
A really frightening joke ...  :ninja:
BTW;
Name: Sayan Mukherjee
Age : 13
From: Kolkata
Nice to meet you.  :P

@potla
Tui terrorist na rock star??? :rotfl:
@franky
Forgot to mention,some girl in our class is a die-hard kasab fan. :rotfl:  :rotfl:
Guess she will believe me if I tell her he's at AoPS.

[quote="Scary math"]Guess she will believe me if I tell her he's at AoPS.[/quote]
Yeah he has some are of 'problem solving' :P Just KILL 'em and the probs over :D

[quote="Scary math"]Forgot to mention,some girl in our class is a die-hard kasab fan. :rotfl:  :rotfl:
Guess she will believe me if I tell her he's at AoPS.[/quote]

either the girl is mentally ill or dis is a sick joke  :wallbash_red:
ask her if she wud feel the same when she wud have the barrel of a fully loaded ak47 shoved up her pu*sy  :ninja:

i always had believed commies are anti-nationals  :P

@Sagar
Well,for ur info,I am the only one what u call by anti nationalist in my class, :P
Earlier,her heartthrob was Shoaib Malik,till a few months ago,it was Kasab,I am not sure of the current one. :rotfl:  :D

dat girl really is sick
EDIT: abusive language

@sagar
Strange the mod hasnt kicked u out yet :mad:
Better stop using such words,and she isnt sick,she is too childish. :furious:

@Scary math I was just viewing the thread.I totally agree with you.Moderator should take action.It can not be tolerated in mathlinks.

ok sorry  :(
i just lost my temper when i read that the girl has a crush on kasab
i still refuse to believe it and i think scary is saying such things just to stand out in the crowd :P

[hide]
edited one post, cant edit the previous one
mods please delete it  :roll:
[/hide]

Man,I am serious! :mad:
Why would I be makin it up?
Better appreciate the fact that some wriedos do have soft corners for these brainwashed,heartless creatures. :P  :rotfl:

What is their to appreciate in it?Its actually funny that such weird creatures exist in India :rotfl:

Well,dat doesnt make her eligible for abusive language :mad:

V^3 realised India forum is a spammer's paradise,he has stopped visiting his own intro!

chill guys....
sagar whats your email id?

@frank
vaze.sagar at gmail dot com
sagarvaze at rediff dot com


SIGNUPS:

1. Joe10112
2. r15s11z55y89w21
3. dragon96
4. BOGTRO
5. vahalla
6. ernie
7. auroraak
8. bluecarneal
9. Bugi
10.mz94

RULES:

You should read each part of the story. Usually, there will be some puzzle. Your goal is to a) figure out the puzzle and b) somehow over throw the school system. You may do any action you could do with the materials provided.

Your stuff: (per person)

10 pencils
24 color pencils
1 protractor
1 pair of scissors
1 ruler
5 rubber bands (many more on teacher's desks, hint hint).
1 box of 100 paper clips

STORY:

[color=red][u][b][size=150]Chapter 1: Windows[/size][/b][/u][/color]

Grimly, the students file in, one by one, into their seats. As they sit down, they see the dark black lens of the security cameras focusing on them. Backs rigid, faces blank, they attempt to pay attention to the dull, flat voice of the teacher. Today, one of the students dropped off. Soft snores came from his nose. Turning around, the teacher swiftly pressed a button, causing the floor beneath the child to disappear. He fell into the hole, never to be seen again...

After the incident, the children paid better attention. At the ring of a computerized bell, the class quietly shuffled off to their next class. As they routinely filled the classroom, Joe10112 was sent to the office to make copies for the class. As Joe10112 quickly stepped through the quiet hallway, he noticed a series of windows that he had always seen covered by shades. Today however, was different. Joe10112 cautiously peered into the window. Robots! Lots and lots of robots. They strode around the room. Some peered into monitors, others worked on a strange machine. As Joe10112 continued to stare into the window, he felt a strange presence. Something was wrong here. He turned around, and everything was black.

All he heard was muffled voices. Straining to open his eyes, Joe10112 finally regained vision. Above him were many concerned faces. r15s11z55y89w21 pulled him up. Joe10112 turned around and saw the windows, except this time, they were closed...

Classroom 3A1Q, chemistry. Once again, the kids were not paying attention. Joe10112’s incident led them all to think about everything except chemistry. Then, an idea sparked. Soon a note spread around the class, and on it, one word. Rebel.

[color=red][u][b][size=150]Chapter 2: Textbook.[/size][/b][/u][/color]

The teacher had them crack open their thick textbooks. As dragon96 did so, a note slipped out on to his desk.

[quote]Puzzle 1:
brmyd[/quote]

The students looked at each other, faces solemn. The question is. What do you do?

Yes, sorry. I wrote most of the story before hand and I must have forgot to include the name.

Isn't that copying off the Race to survival??
Crawl through them to get out, or to that room. I have no Idea what you are trying to point to.

Vents are infamous in all situations.

I'm trying to point to the fact that you do something like:

I use a ruler to pry of the tiled upper celing and climb through the vents.

I use a ruler to pry of the tiled upper celing and climb through the vents. :D

*tells ppl wat hapnd to me*

once in a good place with NO TEACHERS snooping

where did code drop from?

Read the story.

Aha! dragon96 exclaimed. Excitedly, dragon96 used a ruler to pry off the celing panels. Using a desk he reached up to climb up into the vent. As he dragged himself through, he met another code, scratched on to the side of the vent:

[quote]Puzzle 2:
UPJ ZY OPDK[/quote]

Can anyone remind me what Caesarean Cipher arrangement is? I suspect it is, because I have tried other codes, which don't seem to work. Or better yet, tell me if Caesarean works.

Wait, so are we in the hall or the room?

Where do you want to be?
(By the way: I'm going to ask frustrating rhetorical questions from now on).

I think this is too hard. Cause what are you supposed to do?

Decode the message, obviously...if you bothered to read anything?

*sigh*

dojo, EASIER problms plz?

thnks

and is the thread dead????

dojo, post then we know the thread is not dead


geometry
calculus
function
probability
algebra
linear equation

link appear dead

You better post them here

Some server problem please see http://thepizone.com/gi/isi-2009.pdf

[hide="1"]

 We have 3 linear equations:

$ \alpha x \plus{} \beta y \plus{} \gamma z \equal{} 0$
$ \frac {\alpha}{x} \plus{} \frac {\beta}{y} \plus{} \frac {\gamma}{z} \equal{} 0$ , which follows from taking conjugate of the above equation and using |$ \alpha$| =1 .

and $ x \plus{} y \plus{} z \equal{} 0$

This forms a homogeneous equation in x ,y and z. But since x , y , z are non zero , this system has infinite solutions. Hence the determinant of coefficients of the above 3 linear equation should be 0 which gives the required condition.[/hide]

Will post others in evening.

@ Manjil : for the 6th Q, it should be [b]ad – bc not equal to zero[/b],

[b]not ab – cd[/b].

Yes thanks, and question 9 should have been $ \mid f(x)\mid <\frac{3}{2} \forall x \in [\minus{}1,1]$ and not as mentioned.

[hide="Solution to the 5th one"]
$ \frac{10^r}{p} \minus{} \frac{1}{p} \equal{} a_1a_2 \cdots a_r .0$

$ 10^{r}$ [b] –   1 = A.p [/b]  where A = $ a_1 a_2 \ldots a_r$ = a natural no.

$ 10^r  \equiv 1 \ (mod p)$

Q.E.D !
[/hide]

[u]The correct 6th Question[/u] :

If ad - bc is a real number not equal to zero and if $ b_1$ and $ b_2$ are integer multiples of (ad - bc)
then prove that there exists two integers x and y such that ax + by = $ b_1$ and cx + dy = $ b_2$ are both simultaneously satisfied.

[hide="Solution"]

Let $ b_1 \equal{} n_1$ ( ad – bc )

and $ b_2 \equal{} n_2$ ( ad – bc )

where $ n_1$ & $ n_2 \ \in \ I$

ax + by  =  $ b_1$  …. (1)

cx + dy = $ b_2$    …. (2)

eqn (1) x d – eqn (2) x b gives u

 x ( ad – bc ) = d $ b_1$ – b $ b_2$

or  [b] x =  d $ n_1$ – b $ n_2$

which is an integer. [/b]

Similar proof for y .

[i]But stupid me - couldn't think of this in the exam hall [/i] :(

 [/hide]

[quote="Nora.91"][hide="The 4th one"]

Answer was  x = 5/13, y = -12/13,
Min value of x^2 + y^2 being 1.

2 methods to do this : by Geometry and Calculus.

I did it by Calculus taking x = 14 cos@ - 5 and y = 14 sin@ + 12.

[/hide][/quote
isnot   x^2+y^2 minimum when x=y......

[quote="Nora.91"][hide="the 10th one"]

Rational root   =>   D must be perfect square

D = b^2 – 4ac = odd  { odd – even }

And every odd perfect square is of the form $ 8n \plus{} 1 \ ; \ n \in N$

 This gives  8 m + 1 – 4 x ac  =  8 n + 1

or   ac = odd = 2 ( m – n )
which is a contradiction.

So, no rational roots.

[b]
OR  [/b] :idea:

u can prove that the discriminant is of the form [b]8 n + 5 [/b]
so it can’t be the perfect square of an odd integer.

[/hide][/quote]

We can do this by assuming that there is a rational root and coming to a contradiction. This is a very old Canadian math Olympiad problem.

[quote="manjil"]The subjective questions of this year's ISI paper can be found at http://www.googolplexideas.com/index.php/math-olympiad/4-problems/48-isi-2009.html[/quote]

This is now working!

[quote="UKO"]
is not   x^2 + y^2 minimum when x=y......
[/quote]

yes, but this case u also have d condition $ (x\plus{}5)^2 \plus{} (y\minus{}12)^2 \equal{} 14^2$

if x = y, this is not satisfied !!

[hide="2nd first part"]

Let us consider the function $ f(x) = x(x+1)..(x+2009)$

Let $ \alpha$ be a root to the equation in the question, then $ f(\alpha) = c$

Assume $ \alpha$ is a root of multiplicity $ 2$. Then $ f'(\alpha) =c$

Its easy to see that $ f'(x) = f(x)(\frac{1}{x} + \frac{1}{x+1} + .... +\frac{1}{x+2009})$

Put $ x = \alpha$ in the above equation, we get

$ c = c(\frac{1}{\alpha} + \frac{1}{\alpha +1} ....)$

Hence, for the root to have multiplicity atleast $ 2$ we should have $ 1 = \frac{1}{\alpha} + \frac{1}{\alpha +1} ....+\frac{1}{\alpha +2009}$

Now if the root is of multiplicity $ 3$, $ f''(\alpha) = c$

$ f''(x) = f'(x)(\frac{1}{x} + \frac{1}{x+1} + .... +\frac{1}{x+2009}) - f(x)(\frac{1}{x^2} + \frac{1}{(x+1)^2} + ...... +\frac{1}{(x+2009)^2}$

Now putting $ x=\alpha$ and using $ f(x) = f'(x) = c$ and $ 1 = ({\frac{1}{\alpha} + \frac{1}{\alpha +1} ....)}$ , we get

$ f''(\alpha) = c - c(\frac{1}{\alpha^2} + \frac{1}{(\alpha+1)^2} + ...... +\frac{1}{(\alpha+2009)^2})$[/hide]

which cannot be equal to c for any real value of $ \alpha$. Hence proved.

[quote="preetham"]conjurer

shudnt f'(x)=0 ????[/quote]

Incredibly wronged..bare with me please.hehe.

Still continuing with prev solution. we have f'(a) =0

f''(a) = -f(a)(1/a^2 + 1/(a+1)^2 + ... +1/(a+2009)^2) which cannot be 0.

[quote="Nora.91"][quote="UKO"]
is not   x^2 + y^2 minimum when x=y......
[/quote]

yes, but this case u also have d condition $ (x \plus{} 5)^2 \plus{} (y \minus{} 12)^2 \equal{} 14^2$

if x = y, this is not satisfied !![/quote]

Sorry I couldnot understand................... :(
Solving..........(x+5)^2+(x-12)^2=14^2will give us the solution.

Pardon me i wrote a wrong sttment there .... about it not being satisfied

BUT

[quote="UKO"]
is not   x^2+y^2 minimum when x=y......
[/quote]

i don't get this

consider

$ F \equal{} x^2 \plus{} y^2 \\ if \ x \equal{} y \\ then \ F \equal{}  2x^2 \\ but \ if \ y < x \\ then \ y^2 < x^2 \\  \\ So, \  F \equal{} x^2 \plus{} y^2 < x^2 \plus{} x^2 < 2x^2 $

[b]this value is less than when x = y[/b]

so, Why do you say that minimum value is when x = y  :?:

[url]http://targetiit.com/iit_jee_forum/posts/isi_2009_subjective_mib_8978.html[/url]

all the Q have been solved here

[quote="manjil"][quote="Nora.91"]ISI  B.Stat results out ....[/quote]

please give the link and what abt B.math?[/quote]

OK I found the link myself...http://www.isical.ac.in/~deanweb/BSTATINTLIST09.pdf

[quote="rituraj007"]well,y do u never post ur own result :?:

best of luck though..[/quote]

Thanx ...

That's becoz my course results ( B.Math ) aren't out yet !  :rotfl:

and even if they r, its kinda shameful to post it in front of people of your level  :(

@ manjil : i don't know when B.Math will be out, i think we have to check their website everyday.

[quote="apratimgtr"]Hi .

ISI BMath Results are out http://www.isibang.ac.in/~statmath/bmath09.htm[/quote]

I got thr.....Sl.No 60

nora.91 too!

congrats to all who are selected :)

[hide="by the way"]what is diff between B.math and B.Stat(stat is a part of maths,i suppose)[/hide]

@skand : Statistics is a completely different course .... look at their syllabi

[b]Congrats Manjil !!![/b]   :thumbup:

[quote="Nora.91"]

[b]Congrats Manjil !!![/b]   :thumbup:[/quote]

Thanks same to you too!!!!:thumbup:

Are the results make you elligible for interveiw or admission?

i got through too!!! but what worries me now is the interview! do u have any idea what sort of questions are asked in the interview?

[quote="Scary math"]How many are selected for CMI through the written exams?And how many are eliminated through interviews?
Is it true that INMO awardees get direct admission to CMI for BSc?Reply at mahadgos@yahoo.com as soon as possible! :juggle:[/quote]

there are no interview for CMI I guess, abt 45 are selected and yes INMO awardees get direct admsn.

Well my neighbour who appeared for CMI this year said there were interviews.

Yeah, i think there is interview fr CMI

and INMO and INAO awardees get admission only after the interview, they just don't have to take the test.

EDIT : The [url=http://www.isibang.ac.in/~statmath/bmath09.htm]ISI interview schedule's[/url] been put up,

mine on the first day itself !  :noo:

Where could I get the ISI B-Stat intreview results?
 And one more request,I am a 10th grade Math Olympiad begginer therefore I seek some advice:
What is the minnimum qualifying marks required and the minnimum no of questions one should get right in RMO? Which is more important,probability or combinatorics?From where do more questions come? What are the prime topics ?Could I skip convergency & divergency,limits,Recurring series and calculus when preparing for Regional Olympiads?
Please reply at mahadgos@yahoo.com as soon as possible
Yaar koi to help kaar dimag ka  :idea:  jalane mein! :fool:
 :maybe:


Hello friends,
                         Recently I read how dynamic programming can be applied to multiply a chain of matrices (which of course are multiplication compatible) in a reasonable amount of time. What I don’t understand however is why can’t we apply greedy approach for the same wherein we do the following:-

First multiply the matrices M(i) and M(i+1) that minimize the product d(i-1).d(i).d(i+1) and continue in the same way.

We assume that we have a chain of matrices M1.M2.M3-------.Mn-1.Mn  and in addition to that an array d[0…n]  where Mi is of dimension d(i-1).d(i).


Some one of you can help me to find
  "A Classical Introduction to Modern Number Theory"

m.candales, you can find many good books in webpage http://gigapedia.org  [url=http://www.mediafire.com/?3sxz5kg5oyy]here[/url] may a book you needed.

Michael.

I think the book in concern is 'A Classical introduction to Modern Number Theory' by Ireland and Rosen. It is under copyright so please do not try and acquire it illegally.


parameterization
geometry
geometric transformation
rotation
function
quadratics
ceiling function

Hi folks.

I've got a problem here that I am reaching out for help on.  I've gone through 3 courses of calc a few years back, and haven't used it much since.  I've got this problem I'm working on.  I've got an approximation I'll probably use, but my curiosity is pushing me to find a precise equation.

It's relatively simple.  A cloth is wound up on a roll using a motor rotating at a constant speed.  I'd like to increment the cloth by a constant length for the entire roll.  I suppose the explanation of it is that simple.

Not that I expect anyone to give me the answer, but in case you'd like to know, the parameters of the system are:

Empty Core Diameter: 1.95 inches
Full Cloth Diameter: 4.75 inches
Motor RPM: 0.89
Total Cloth Length: 75 feet
Desired Increment: 0.5 inch

I can calculate that:
Empty cloth speed: 0.091 inch/sec
Full Cloth Speed: 0.221 inch/sec
Motor On Time (Empty): 5.505 sec
Motor On Time (Full): 2.26 sec

I'll probably end up using a linear approximation, but I'd love to know the precise equation.

Can anyone give me a kick start?

With this design, the speed at which the cloth is added is proportional to the radius of the spool. There are limits; the motor is working a lot harder with a full roll.

Let $ C(t)$ be the amount of cloth, and $ r(t)$ be the radius. We have $ C(t)\equal{}a(r(t)^2\minus{}r_0^2)$, where $ a$ is an empirically derived constant and $ r_0$ is the initial radius; this just estimates $ C$ by the volume taken up. In addition, $ C'(t)\equal{}br(t)$, where $ b$ is another constant (found either empirically or from the rotation rate).

Now, we can also differentiate our equation for $ C$, and get $ C'(t)\equal{}ar(t)r'(t)$. This implies that $ b\equal{}ar'(t)$, and $ r(t)\equal{}r_0\plus{}\frac{b}{a}t$.
From that, we can find $ C(t)\equal{}\frac{b^2}{a}t^2\plus{}2br_0t$.

The exact increment of $ t$ needed to increase $ C$ by a certain amount involves a square root.

Wow, great.

I think that makes sense.  I think what was always missing in school was teaching the concepts needed to be able to model a problem like this.  By the time I got done with all the math I took I could manipulate all sorts of equations, but couldn't model the most basic of real life scenarios.

So to recap what you did.  You could write an equation for the volume of the cloth on the take up roll, and know that that is proportional to the length of cloth.

Then you can write an equation stating that the rate of change of the length of cloth was proportional to the take up roll radius.

After that, it's a matter of differentiating the first, equating to the second and solving for the radius as a function of time, and plugging back into the original equation.  Use algebra to solve, and end up with a quadratic equation.  Simple and brilliant.

So I'm left with an equation that gives me the amount of cloth on the roll for a given amount of time the motor has been running.

But I need something slightly different.  I need to calculate a preset for a timer in a PLC to turn on the cloth motor for the Nth time.

So how to take the next step.....  As you stated it requires a square root.  I can set C(t) to some number, then to a number 1/2" greater, and find delta t for each iteration, but that won't work well in a PLC program...

What I really need is an equation for t(n), if that makes any sense.

This is exactly the help I needed to get me started - thanks a lot!

I'm assuming you mean you need the time it takes for your motor to wind $ n$ partitions of some length $ l$?... I really can't go on without knowing: assuming your motor runs at constant power I think the rotational velocity of the spool can be written as a function of time. Once you have that done you can probably use rotational kinematics to find nearly anything you need.

igiul- the motor is running at constant speed, not constant power. It's simpler than that.

jbenderjb- It makes sense- that's the inverse function. Taking the inverse of a quadratic function is essentially solving a quadratic equation:
$ \frac {b^2}at^2 \plus{} 2br_0t \minus{} C \equal{} 0$
$ t \equal{} \frac { \minus{} 2br_0\pm\sqrt {4b^2r_0^2 \plus{} \frac {b^2}{a}C}}{2\frac {b^2}{a}} \equal{} \frac { \minus{} ar_0\pm a\sqrt {r_0^2 \plus{} \frac1aC}}{b}$

We choose the positive sign in the square root; time runs forward. If we put in the $ n$th value of $ C$, we get the $ n$th value of $ t$ back from this formula.
This $ t$ is the total time the motor has spent running.

[quote="jmerry"]igiul- the motor is running at constant speed, not constant power. It's simpler than that.[/quote]

I was actually think of finding the moment of inertia as a function of time. From there, equating $ KE_R$ to the work of the engine allows us to isolate $ w$ (rotational velocity). I think your point is that the engine runs at a constant RPM whereas I took to the notion that this implied that the engine would do the same amount of work winding the cloth albeit it slows down as mass on the spool increases.

Correct me if I'm wrong; it's been a year since I took advanced physics so my terminology and fluency on the subjects at hand are a bit rusty.

It is a simple set-up - as I'd like to define it, at least.

I have a motor that will spin at a constant speed when on.  It is winding up a spool of cloth which cleans something.  Every once in a while I turn the motor on to increment the cloth to get a fresh section of cloth.  I'd like to have the same cloth increment each time over 75 feet of cloth, so I need some decrement logic to take into account the changing take up roll diameter.

I am assuming constant speed for our purposes.  It is, in reality, a constant power system, but through an 1800:1 gearing without all that much friction to be overcome.  So I think the constant speed is a very safe estimate.

This will be programmed into a computer of sorts (PLC).  Working in real world, I need to take into account the fact that my timer precision is in milliseconds, and so in the end I will probably keep track of how long the motor has run so far, and calculate how long it should run for another 1/2 inch increment.  Just to make things more fun, it will be an adjustable increment by the user.

This has been huge help, and I look forward to writing some of this out.

Let the full cloth radius be $ r_f$ and the empty core radius be $ r_o$
$ r_o = 0.975\text{in}$
$ r_f = 2.375\text{in}$
$ \Delta r = 1.40\text{in}$

Then, $ r_n = r_o + n\rho$ where $ \rho$ is the thickness of our cloth. There exists $ n: r_f = r_o + n\rho$ which implies $ n\rho = \Delta r$
[hide="The Tricky Part"]
\begin{eqnarray*}75 & = & 2\pi\sum_{i = 1}^n r_n \\
& = & 2\pi\displaystyle\left(\sum_{i = 1}^n r_o + \rho\sum_{i = 1}^n i\right) \\
& = & 2\pi\left(r_on + \rho\frac {n(n + 1)}{2}\right) \\
& = & \pi n(2r_o + \rho n + \rho) \\
& = & 3.35\pi n + 1.4\pi \\
\end{eqnarray*}
[/hide]
From there I found $ n$, then $ \rho$ by $ \Delta r = n\rho$.

$ n\approx 6.708430288$
$ \rho\approx 0.208692636$

Thus, $ r_n = r_o + \rho\lceil 0.89t\rceil$ where $ \lceil x\rceil$ denotes the least integer greater than or equal to $ x$. And finally, for $ n$ partitions of half inch strips, $ \boxed{t(n,t_e) = .89\pi n(r_o + \rho\lceil 0.89t_e\rceil )}$ where $ t_e$ is total time elapsed. $ t(n)$ [i]should[/i] hold so long as $ r_n$ does not change during the interval $ (t_e,t_e + t(n,t_e)$.

Funnily enough, I used absolutely no calculus at all to do that.

[b]edit:[/b] The radius that $ t(n)$ generates is that on the outside of each layer of cloth, so you may want to alter that just a bit by subtracting $ \frac{\rho}{2}$ from our $ r_n$ function.

OK.  That's a lot for me to think about.

I can get to the point where you define the radius at any given time (accumulated motor run time)

I'm lost at the point where that is multiplied by 0.89 * pi * n   (Sorry I don't know TeX...)
The rotation rate times ( pi * radius ) gives us a "cloth speed" at a given time
Though shouldn't it be multiplied by 2 for the circumference?

So then we multiply by n...  Is n still the total number of rotations in a full cloth?

Conceptually, I can't wrap my head around that one...

The cloth length is actually 75 feet, so 900 inches, but I can easily recalculate the p and n.

I've been thinking about this all day...

I can see where "cloth rate" is divided by the partition length, giving you a "partition rate", and that would account for what I thought was a missing "2" for the circumference.

n is the "partition count", and I just can't figure out why we multiply by that.  (Of-course I could be way off all together)

But if I have partition speed given run time, couldn't I take the inverse of that to give me a "seconds per partition"?

Yes, and that's the linear approximation you were talking about at the beginning. Your increments are short enough that the error in that approximation is negligible.


factorial
AMC
USA(J)MO
USAMO
induction
algebra proposed
algebra

Help me find value of $ \sum_{k\equal{}0}^{n}k!(n\minus{}k)!$ in terms of $ n\in\mathbb N$.

Thanks in advance for your nice solution.

Mathematica gives :
$ \minus{}i 2^{\minus{}(1\plus{}n)} [\pi \minus{}i \beta(2,2\plus{}n,0)] \Gamma(2\plus{}n)$

Since we are in "proposed and own" category, the OP has the solution.

So, after strange mathematica and mapple solutions, could we have J.Y.Choi's one, please ?

I posted this problem in this category because I made this problem and couldn't solve it by myself.

My insignificant approach is thinking of $ \binom{n}{k}^{\minus{}1}$, but I can't make an advance any more. I need mathlinkers' help.

[quote="J.Y.Choi"]I posted this problem in this category because I made this problem and couldn't solve it by myself.[/quote]
Please, read the forum definitions and respect the usage. It's just a matter of politeness :

[i][u]Unsolved category[/u][/i] : "Problems you couldn't solve and to which you know that there is a solution (i.e. a problem from a contest, etc.) but you don't know it."

[i][u]Proposed and own category[/u][/i] : "Problems [u][b]that you have already solved [/b][/u]and you are interested in second opinions or solutions."

[i][u]Open category[/u][/i] : "An open question is a question that has no known solution up to this moment, and it is not known wheter the problem has one or not."

Some users are interested in only some categories (personnaly, I generally spend more time on "proposed and own" than on "Open" since I'm sure there is a clever answer (olympiad level) and I know, when I fail finding it, that original poster will post the answer when asked.

In your case : you randomly created a problem and dont know if there is a solution : Post it in "Open" category and write in the beginning that you created it.

If my memory serves me right this is an adaptation of a USAMO problem. Its mentioned somewhere in Titu's book: Mathematical Olympiad Challenges

The given series is the form of $ n!\sum_{k\equal{}0}^{n}\binom{n}{k}^{\minus{}1}$. So,let's denote a sequence $ a_n$ by $ a_n\equal{}\sum_{k\equal{}0}^{n}\binom{n}{k}^{\minus{}1}$. For $ n\geq 2$, we have

$ a_n\equal{}\sum_{k\equal{}0}^{n}\frac{1}{\binom{n}{k}}\equal{}\sum_{k\equal{}0}^{n\minus{}1}\frac{k!(n\minus{}1\minus{}k)!(n\minus{}k)}{(n\minus{}1)!n}\plus{}1$
$ \equal{}\sum_{k\equal{}0}^{n\minus{}1}\frac{1}{\binom{n\minus{}1}{k}}\minus{}\sum_{k\equal{}0}^{n\minus{}1}\frac{k}{n}\frac{(n\minus{}1\minus{}k)!k!}{(n\minus{}1)!}\plus{}1$
$ \equal{}a_{n\minus{}1}\minus{}\sum_{k\equal{}0}^{n\minus{}1}\frac{k}{n}\frac{1}{\binom{n\minus{}1}{k}}\plus{}1\quad\cdots\boxed{1}$.

Moreover, $ \sum_{k\equal{}0}^{n\minus{}1}\frac{k}{n}\frac{1}{\binom{n\minus{}1}{k}}\equal{}\sum_{k\equal{}0}^{n\minus{}1}\frac{n\minus{}1\minus{}k}{n}\frac{1}{\binom{n\minus{}1}{k}}\equal{}\frac{n\minus{}1}{n}a_{n\minus{}1}\minus{}\sum_{k\equal{}0}^{n\minus{}1}\frac{k}{n}\frac{1}{\binom{n\minus{}1}{k}}$.

Therefore $ \sum_{k\equal{}0}^{n\minus{}1}\frac{k}{n}\frac{1}{\binom{n\minus{}1}{k}}\equal{}\frac{n\minus{}1}{2n}a_{n\minus{}1}$.

Finally, by $ \boxed{1}$, we get $ a_n\equal{}a_{n\minus{}1}\minus{}\frac{n\minus{}1}{2n}\plus{}1\equal{}\frac{n\plus{}1}{2n}a_{n\minus{}1}\plus{}1$. Would anyone can solve this inductive form for $ a_n$? I have no idea...

I can't find it in Mathematical Olympiad Challenges, but Kurlyandchik and Lisizkiy prove in Kvant 10/1978 that $ \sum_{k\equal{}0}^n \frac{1}{\binom{n}{k}} \equal{} \frac{n\plus{}1}{2^{n\plus{}1}}\sum_{k\equal{}1}^{n\plus{}1}\frac{2^k}{k}$. Of course, the proof is trivial by induction since J.Y.Choi has done all the hard work. I doubt that the term on the right hand side can be simplified.

  darij


calculus
integration
function
trigonometry
calculus computations

Find the function $f(x)$ such that $f(x)=\cos (2mx)+\int_{0}^{\pi}f(t)|\cos t|\ dt$ for positive inetger $m.$

The integral equality implies that

$f(x)=\cos(2mx)+C$ where $C$ is a constant.

Fpr calculating $C$ use the integral equality and you should get that

$C=\int_{0}^{\pi}(C+\cos(2mt))|\cos(t)|dt$

$C=C\int_{0}^{\pi}|\cos t|dt+\int_{0}^{\pi}\cos(2mt)|\cos(t)|dt$.

Hence, $C=\frac{-1}{3}\int_{0}^{\pi}\cos(2mt)|\cos t|dt$.


Find all polynomial with real-valued coefficients $P$ and $Q$ satisfying
(a) $2P(x)-Q(x)=P(y)-y$ ,
(b) $P(x)Q(x)\geq x+1$ ,
for all real numbers $x$ and $y$

This one is really easy... :?  Why is it still unsolved?

Pierre.

Ok, so I will do it :
Since the lhs of the first condition does not depend on $y$, it follows that the rhs is constant, so that $P(y)=y+c$ for some constant $c$.
From the first condition, it now follows that $Q(x)=2x+c.$

Thus the second condition asks for $2x^2+3cx+c^2 \geq x+1$ for all $x$
which means that $2x^2 + (3c-1)x + c^2 - 1 \geq 0.$
This is equivalent to ask that the discriminant of this quadratic is non-positive.
But $\Delta = (c-3)^2$ thus $c=3$ which gives $P(x) = y+3$ and $Q(x) = 2x+3.$

Pierre.

Are P and Q functions there?  I just don't understand your notation...

P and Q are polynomials with real coefficients, as the question says.

Ah...it appeared that P and Q were the coefficients...


geometry
3D geometry
algebra
polynomial
Rational Root Theorem

Rationalize the denominator.

$1$ divided by the square root of $(1+\sqrt{2})$

Edit: I don't know whats wrong. Can someone help me?
Edit2: okay i fixed it. kinda.

[quote="anyone"]you mean $\frac{1}{\sqrt{1+\sqrt{2}}}$ ??[/quote]
yea. I didn't know until now.

sooo... does anyone know how to do it?  :lol:

Let the equation = x, then take x^2, multiply num. and denom. by 1-sqrt(2), then take the square root again:

sqrt(sqrt(2)-1)

lol, nice, that's definitely better than my method.

[quote="davidyko"]Mm, this one gets a bit ugly.
But basically you multiply both the numerator and the denominator first by $\sqrt{1+\sqrt{2}}$, and then by $1-\sqrt{2}$, and simplify.
Then, you would get $\sqrt{2+2\sqrt{2}}-\sqrt{1+\sqrt{2}}.$[/quote]
Thats exactly what I got. I just wasn't sure.

[quote="now a ranger"][quote="davidyko"]Mm, this one gets a bit ugly.
But basically you multiply both the numerator and the denominator first by $\sqrt{1+\sqrt{2}}$, and then by $1-\sqrt{2}$[/quote]

The book said to multiply, but I don't see how you multiply something with two brackets above it. I always thought of it as a cube root. Could someone elaborate? how to mutiply multiple brackets like that. It really confused me.[/quote]

You mean an embedded radical? The square root of a square root is a fourth root, by the way. (You can see this by writing the square roots as exponents.)
If you're multiplying two radicals together, and one or both have embedded radicals, don't panic; just multiply the stuff inside the outer square roots together as if it weren't there, and then just slap the square root on top when you're done.

For example,
$(\sqrt{2})({\sqrt{1+\sqrt{2}})=\sqrt{(2)(1+\sqrt{2})}=\sqrt{2+2\sqrt{2}}}$.

Off-topic: Oh, cool, post #250. Do I have two-and-a-half stars now...? Let's see.

woo! nice method.

The bracket there kept driving me nuts.

Is it because the 2 brackets make it the square root of 1 + the 4th root of 2. In the p+q part, there has to be a bracket around it, or the number is the square of the one with the brackets.

[quote="now a ranger"]the 2 brackets make it the square root of 1 + the 4th root of 2. [/quote]

Clearly false.  By this logic, $\sqrt{4+4}= \sqrt{4}+\sqrt{4}= 4$, but of course $\sqrt{4+4}= 2 \sqrt{2}$.

oh now I see what you meant by "showing things". It says to "show" that something is true or false in aops 1. Ooo  :blush:

[quote="t0rajir0u"][b]Follow-up:[/b]  Show that $\sqrt{1+\sqrt{2}}$ cannot be written in the form

$p+q \sqrt{2}$

Where $p, q$ are rational.[/quote]

[hide="Ugly Solution"]
Proceed by contradiction:
$\sqrt{1+\sqrt{2}}=p+q \sqrt{2}$
$1+\sqrt{2}= p^{2}+2pq\sqrt{2}+2q^{2}$

So equation rational and irrational parts:

$1= p^{2}+2q^{2}$
AND
$1\sqrt{2}=2pq\sqrt{2}$

Simplifying:
$1= p^{2}+2q^{2}$
AND
$pq=\frac{1}{2}$

So $p = \frac{1}{2q}$

Substitute that into the first equation:

$1= (\frac{1}{2q})^{2}+2q^{2}$
$1 = \frac{1}{4q^{2}}+2q^{2}$
$4q^{2}=1+8q^{4}$
$8q^{4}-4q^{2}+1 = 0$
By rational root theorem, there are no rational roots of this equation.  Hence, $\sqrt{1+\sqrt{2}}$ cannot be written in the form $p+q \sqrt{2}$ where $p, q$ are rational.

EDIT: Actually, the discriminant is negative, so it doesn't even have any REAL solutions.  (I think).

[/hide]

Contradiction is indeed a good way to go, but not quite in that form.

[hide="Instead take this route..."] Yes, we have $1 = p^{2}+2q^{2}$ and $1 = 2pq$.  Write this as

$p^{2}+2q^{2}= 2pq$

Now, clear the denominators and let $p = \frac{a}{d}, q = \frac{b}{d}$, where $gcd(a, b, d) = 1$ are integers.  Then we wish to solve

$a^{2}+2b^{2}= 2ab$

In integers.  Well, $2 | a$, so let $a = 2c$.  But then

$2c^{2}+b^{2}= 2bc$

Which implies no solutions by infinite descent.  QED. [/hide]
Edit:  [hide="Alternately..."] If $p^{2}+2q^{2}= 2pq$ then

$(p-q)^{2}=-q^{2}$

Which clearly has no nontrivial solutions. [/hide]


Daca ABC este un triunghi ascutitunghic, sa se afle maximul expresiei:
\[ E \equal{} \frac {\cos A\cos B\cos C}{\sin ^{2} A\sin ^{2} B\sin ^{2} C} .
\]

$ \begin{array}{l}
 tgAtgBtgC \equal{} tgA \plus{} tgB \plus{} tgC; \\
 \frac{1}{{(tgA \plus{} tgB \plus{} tgC)\sin A\sin B\sin C}}\mathop  \le \limits^{Jensen} \frac{1}{{3\sqrt 3 \sin A\sin B\sin C}}(*); \\
 \sin A\sin B\sin C \equal{} \frac{1}{2}(\cos (A \minus{} B) \minus{} \cos (A \plus{} B))\cos C; \\
 \left. \begin{array}{l}
 \cos (A \minus{} B)\cos C \equal{} \frac{1}{2}(\cos (A \minus{} B \plus{} C) \plus{} \cos (A \minus{} B \minus{} C))(1) \\
 \cos (A \plus{} B)\cos C \equal{} \frac{1}{2}(\cos \pi  \plus{} \cos (A \plus{} B \minus{} C))(2) \\
 \end{array} \right\| \Rightarrow \sin A\sin B\sin C \equal{}  \\
 \frac{1}{4}((\cos (A \minus{} B \plus{} C) \plus{} \cos (A \minus{} B \minus{} C)) \minus{} (\cos \pi  \plus{} \cos (A \plus{} B \minus{} C))) \equal{} \frac{1}{4}(\cos (\pi  \minus{} 2B) \plus{}  \\
 \cos (\pi  \minus{} 2A) \plus{} 1 \minus{} \cos (\pi  \minus{} 2C)); \\
  \minus{} (\cos 2B \plus{} \cos 2A \plus{} \cos (\pi  \minus{} 2C))\mathop  \ge \limits^{Jensen}  \minus{} 3\cos \frac{{2B \plus{} 2A \plus{} \pi  \minus{} 2C}}{3} \equal{}  \minus{} 3\cos (\pi  \minus{} \frac{{4C}}{3}); \\
 \sin A\sin B\sin C \ge  \minus{} \frac{3}{4}\cos (\pi  \minus{} \frac{{4C}}{3}) \ge  \minus{} \frac{3}{4} \\
 \end{array}$ pentru $ C\equal{}\frac{{{\rm 3}\pi }}{{\rm 4}} \frac{1}{{3\sqrt 3 \sin A\sin B\sin C}} \le \frac{{ \minus{} 4}}{{9\sqrt 3 }}$ dar egalitate nu se obtine in inegalitatea a doua in sensul lui Jensen deoarece nu avem 2B=2A=$ \pi \minus{}2C$


Έστω τρεις κύκλοι, ω1, ω2 και ω3 με κέντρα Ο1, Ο2 και Ο3 αντίστοιχα, που εφάπτονται ανά δυο. P1 και P2 τα σημεία επαφής των ω3 με ω1 και ω3 με ω2 αντίστοιχα. ΑΒ μία διάμετρος του ω3 Οι ΑΡ1 και ΒΡ2 τέμνουν τους ω1 και ω2 στα Χ και Υ αντίστοιχα. Αν Ζ το σημείο τομής των ΑΡ2 και ΒΡ1, ν.δ.ο. Χ,Υ,Ζ συνευθειακά.

[quote="kalantzis"]Πολύ καλή η λύση σας και πιο απλή από αυτή που σκεφτόμουν. Εγώ απέδειξα ότι τα $ P_1,P_2,P_3,Z$ είναι ομοκυκλικά και με βάση αυτό τα σημεία $ Y,Z,P_3$ και $ X,Z,P_3$ είναι συνευθειακά άρα ισχύει και το ζητούμενο.

Κάτι ακόμα. Το σχήμα αυτό πως το φτιάξατε? :)[/quote]
Το σχήμα με το Word


I though it would be cool if everyone who is graduating this year posts what college they plan on attending.  People in college now can post their college also.

This is an excellent thread for this time of year.

Just for completeness, I will note that my undergraduate university was the University of Minnesota College of Liberal Arts, the only college to which I applied. I was in class of '80 there, majoring in Chinese language.

[url=http://www.mtu.edu/]Michigan Technological University[/url], where I will be getting so many AP credits that I will likely graduate early with a double major. I'll be there for grad school as well (hey, they're going to pay me to go there for grad school. I'm not looking further than that)

[quote="ComplexZeta"]I think most grad schools will pay you to go there anyway.[/quote]
Really? MTU offered full tuition, partial fees, and a monthly stipend (oh, and guaranteed admission if I keep a 3.0 GPA in undergrad). Can I get this deal elsewhere too?

A graduate student here told me that at UCSB, if you teach a few classes, then you get free tuition and a stipend of something like $15000 a year, if I recall correctly. I don't think you can get guaranteed admission, but if you do well, you will presumably have many options for graduate school (or at least I hope so).

Princeton University, Simon Fraser University, or the University of Waterloo.

Hmm, it seems Princeton is quickly becoming over-represented :)

I attend Harvard.

[quote="JBL"]Hmm, it seems Princeton is quickly becoming over-represented :)
[/quote]

That's never possible! Go Tigers!


[quote=mohan77]doesn't "simplified fraction" mean the answer it 1 3/4[/quote]

25/16 is not equal to 1 3/4. Where in the problem does it say simplified fraction anyway?

[quote=mohan77]doesn't "simplified fraction" mean the answer it 1 3/4[/quote]

$1\dfrac{3}{4}\neq\dfrac{25}{16}$
The problem just says:
$ \cfrac{\cfrac{3}{8}+\cfrac{7}{8}}{\cfrac{4}{5}}=$