number theory
prime numbers
number theory proposed

Let $a,b$ are  arbitrary positive integers. If $p,q$ are prime numbers prove that there is $s$ prime ,such that ${s}\mid{a}^{pq}-{b}^{pq}$ and ${s}\equiv{1}{(modpq)}$

And it's put in the competition of Mongolia.


geometry
analytic geometry
Pythagorean Theorem

Hope this is the right level.

Fold an 8" by 11" piece of paper so that the upper left hand corner is on top of the lower right hand corner. How long is the fold?

I tried to solve it by using the Pythagorean Theorem and some algebra, but I am not sure if my answer is correct.Thanks in advance.

Well, put the paper on the coordinate axis with 8 on the x-axis and 11 on the y-axis, with the lower left hand corner on (0,0). Then the upper right-hand corner is at $ (8,11)$.

Now, the equation of the line going through $ (0,0)$ and $ (8,11)$ is $ y\equal{}\frac{11}{8}x$

The equation of the fold line is the perpendicular to this line going through it's midpoint.

This line has equation $ \minus{}\frac{8}{11}x\plus{}\frac{32}{11}\plus{}\frac{11}{2}$

Now, just find the points of intersection, blah,blah,blah.

The fold is perpendicular to a diagonal of the page. The length of the diagonal is $ \sqrt {\left(\frac {17}{2} \right) ^2 \plus{} 11^2} \equal{} \frac {\sqrt {773}}{2}$. By similar right-triangles, the fold line will be $ \frac {17/2}{11}$ times as long, or $ \boxed{\frac {17}{44} \sqrt {773} \; \; \text{in}}$, or about $ 10.72 \; \text{in}$.


Find all integers $n$ such that $2^8+2^{11}+2^n$ is a perfect square.

[hide]
Since $(2^4+2^6)^2=2^8+2^{11}+2^{12}$, $12$ is one possible answer.

Now we have to check and see if that is the only possible answer.

Let $2^8+2^{11}+2^{n}=x^2$.
Then $9\cdot2^8+2^n=x^2$, so $2^n=x^2-(3\cdot 2^4)^2=(x-48)(x+48)$.
$x-48$ and $x+48$ both have to be powers of 2.
Since the only power of 2 that differ by 96 are $32=2^5$ and $128=2^7$, the only possible case is $2^n=2^5\cdot 2^7=2^{12}$, so $n=\boxed{12}$[/hide]

frt - there is a typo which you may   like to edit, you mean $x^2$ and not $2^x$

another solution, factoring out the $2^{8}$,
$2^8(1 + 2^{3} + 2^{n-8}) = k^{2}$
$2^4\sqrt{(1 + 2^{3} + 2^{n-8})} = k$ for it to be a perfect square, the radical has to be reducable
$(a+1)^{2} = a^2 + 2a +1 = 1 + 2^{3} + 2^{n-8}$
$a^2 + 2a = 2^{3} + 2^{n-8}$ there are 2 ways to compare terms,
$2a=2^{3}$, $a=4$, $2^{4} = 2^{n-8}$ $n = 12$
or
$a^2 = 2^{3}$, $a = 2^{\frac{3}{2}}$, but that would make it not a perfect square since there would be a radical, therefore we have considered all possibilities, and the only answer is $\boxed{n=12}$


[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
$ \sum_{cyc}\sqrt{a^{2}\plus{}ab\plus{}b^{2}}\geq\sum_{cyc}\sqrt{2a^{2}\plus{}ab}.$[/quote]

Very nice problem  :)  .
WE know that $ \sqrt{a^{2}\plus{}ab\plus{}b^{2}}\geq\frac{\sqrt{3}}{2}(a\plus{}b)$ .
So $ \sum_{cyc}\sqrt{a^{2}\plus{}ab\plus{}b^{2}}\geq\sqrt{3}(a\plus{}b\plus{}c)$

So we have to show that $ \sqrt{3}(a\plus{}b\plus{}c)\geq\sum_{cyc}\sqrt{2a^{2}\plus{}ab}$  or
$ a\plus{}b\plus{}c\geq\sum_{cyc}\sqrt{a\frac{2a\plus{}b}{3}}$

But by AM_GM we have $ \sqrt{a\frac{2a\plus{}b}{3}}\leq\frac{a\plus{}\frac{2a\plus{}b}{3}}{2}$
Summing up the desired  :wink:


Hello everybody,
                              Well friends I had a strange chance of going through Martin Gardner’s Mathematical Puzzles and diversions. The problem that follows is from that book only.
(In my words)

Five men and a monkey are trapped in an island. They collect as many coconuts as they can (to ensure survival in an isolated island). When night dawns and all of them go to sleep, one man secretly wakes up, tiptoes to the coconut basket, divides the coconut into 5 equal parts but finds that such division leaves one coconut aside. He gives that coconut to the monkey and steals one-fifth of the remaining. He then goes back to sleep. After sometime the 2nd person wakes up and does the same thing with the coconuts that remain i.e. he also divides the coconuts into 5 equal parts and finds that one coconut is left out which he hands over to the monkey, steals the one-fifth and goes back to sleep. Similar kind of work is done by all the persons and consequently monkey gets 5 coconuts in the night itself. In the morning they decide to divide the coconuts (remaining coconuts of course). Nobody frets the absence of coconuts because everyone is as guilty as the other one. After division into 5 equal parts they again find that one coconut is left aside. They hand it to the monkey and thus, give him 6 coconuts in all.

Now find the minimum number of coconuts that were there in the basket.

Below I am presenting Gardner’s solution (I don’t want to rob anyone of the joy of problem solving. I present this just for the sake of completing what I read in the book. [u][b]Other solutions are invited[/b][/u])

Solution:-

First observe that if N is any solution (not necessarily the minimal solution), so is N+5^6.
This is evident because of the 5^6 coconuts added none will go to the monkey thereby the fact that it gets 6 coconuts in all is maintained.

So if we allow some P to be our minimal solution, mathematically P+5^6, P+2.5^6 etc are all solutions (not minimal). However, mathematically P-5^6 is also a solution. Sure enough it will be negative.

It is easy to see that one negative solution is -4. To convince yourself consider the following argument. The 1st person gives one +ve coconut to the monkey. Thus he finds
-5 coconuts left. He takes away one-fifth of this and the basket again has -4 coconuts. The 2nd person comes and does the same thing i.e. of the -4 coconuts that he sees he gives one coconut to the monkey and sees -5 coconuts. Thereafter he takes one fifth of this away and the basket again has -4 coconuts. This process continues till the monkey gets 5 coconuts in the night and the basket still contains -4 coconuts. In the morning after giving one coconut to the monkey and thereby finding one -5 coconuts in the basket the 5 men divide it equally among themselves.

Thus it is clear that -4 is indeed the solution to our problem and therefore, the minimal +ve solution to our problem is, no doubt, -4+5^6 = 15621. (Using if N is a solution so is N+5^6).

[b]Hey don't abuse me [/b]for solving the problem right here I was just completing what I read. Other solutions to this problem are cordially invited.

Lol, I remeber doing this problem 2 years ago;  I found a general solution by using matrices (you will find a very nice pattern when you solve it).  Give it a try.

Hi there,
                 Mak, you said that you have a nice solution involving matrices. I would be glad if you will post it here on this site as I was  unable to think of one.

Thanks in advance...


For any positive integer n, let $a_{n}$ denote the integer lying next to $\sqrt{n}$. [In other words, $a_n$ is the integer part of $\sqrt{n}+\frac{1}{2}$.] Compute the sum

\[ \frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+...+\frac{1}{a_{2004}} \]

Darij

Oh yes, this is a bit too intricated...

The correct solution is $\displaystyle \frac {1328} {15} = 88 + \frac {8} {15}$, where the latter expression is more instructive  ;)

  Darij

I understood the problem as $a_n$ is the least integer strictly greater than $ \sqrt {a_n}$...
Thus $a_1 = 2, a_2 = 2, a_3 = 2, a_4 = 3,...$

Pierre.

I'm very sorry, I should not have used the word "next" in this context, it is ambiguous. I meant by "integer next to $\sqrt{n}$" the integer z minimizing the expression $\left| z-\sqrt{n}\right|$ (there cannot be two different z with this property, since square roots of natural numbers are either natural or irrational).

  Darij

Suppose $a_n = k$.  Then $k \leq \sqrt n+\frac12 < k+1$ or after simplification,
$k^2-k+\frac14 \leq n < k^2 + k + \frac14$.  Since $n \in N, k^2-k+1 \leq n \leq k^2 + k$.
Therefore, $a_n = k$ for $k^2 +k - (k^2 - k + 1) + 1 = 2k$ values of $n$.  It follows that $\displaystyle \sum^{2004}_{i=1} \frac1{a_i} = \sum^{44^2+44}_{i=1} \frac1{a_i} + \sum^{2004}_{i = 45^2-45 +1} \frac1{a_i} = 2(44) + \frac{24}{45} = 88 + \frac8{15}$.

Yes, this was exactly my solution.

  Darij

And this is a similar way to solve 'my' problem....

Pierre.


A drawer contains 120 socks. Each sock is one of four colors. What is the minimum number of socks that must be drawn at random to guarantee that two socks of the same color are drawn?

The number of socks in the drawer does not matter. In theory, you could draw 4 socks of different colors. The fifth sock you draw would match one of those first 4, making the answer $ \boxed{5}$.


ABCD is convex quadrilateral with AD = DC = CB and <A + <B = 120.

a) If DN and CN are bisectors of <D and <C, prove that N lies on AB and <AND = <DNC = <BNC = 60
Prove that:
b) AO = OM = BO
c) CO/AO + DO/BO = 1
d) AC < 2BD, BD < 2AC
e) $ S_{ABO} \equal{} S_{ADO} \plus{} S_{BCO}$ ($ S_{ABO}$ is the area ot ABO)
f) $ H_1H_2$ = AB/2
g) DCP is equilateral triangle if and only if ABP is equilateral triangle
h) AO = PO if DCP and ABP are equilateral triangles

$ M \equiv AD \cap BC,$ $ O \equiv AC \cap BD,$ $ H_1, H_2$ are feet of normals from D, C to AB, P is vertex of equilateral triangle on CD or AB.
_______________________________________

[b](a)[/b] $ \angle A \plus{} \angle B \equal{} 120^\circ \Longrightarrow \angle D \plus{} \angle C \equal{} 240^\circ,$ $ \angle CDN \equal{} 180^\circ \minus{} \frac {\angle C \plus{} \angle D}{2} \equal{} 60^\circ.$ The $ \triangle BCD,\ \triangle CDA$ are isosceles with C-, D-angle bisectors $ CN \perp BD,\ DN \perp AC$ $ \Longrightarrow$ BCDN, CDAN are rhombi, $ \angle BNC \equal{} \angle CND \equal{} \angle DNA \equal{} 60^\circ,$ they add to $ 180^\circ,$ A, N, B are collinear.

[b](b)[/b] $ \angle AMB \equal{} 180^\circ \minus{} (\angle A \plus{} \angle B) \equal{} 60^\circ.$ $ \triangle ANC,\ \triangle BND$ are isosceles with $ \angle CNA \equal{} \angle DNB \equal{} 120^\circ$ (rhombi BCNA, BCND) $ \Longrightarrow$ $ \angle NBO \equal{} \angle NAO \equal{} 30^\circ,$ O is circumcenter of the $ \triangle ABM,$ OA = OB = OM = R is the circumradius.

[b](c)[/b] Normal to AB through O cuts arc of the circumcircle (O) of the $ \triangle ABM$ at P, OP = OA = OB = R, AP = BP, $ \angle APB \equal{} \angle AMB \equal{} 60^\circ,$ $ \triangle ABP$ is equilateral. AOC, BOD bisect the $ \angle BOP,\ \angle AOP \Longrightarrow$ BCPO, ADPO are rhombi, CP = CB = CD, DP = DA = CD, the $ \triangle CDP$ is equilateral. $ \angle CPD \plus{} \angle COD \equal{} \angle CPD \plus{} \angle AOB \equal{} 180^\circ,$ CPDO is cyclic. By Ptolemy theorem, $ CO \cdot DP \plus{} DO \cdot CP \equal{} OP \cdot CD.$ Reducing by DP = CP = CD $ \Longrightarrow$ CO + DO = OP = R, $ \frac {CO}{AO} \plus{} \frac {DO}{BO} \equal{} \frac {CO \plus{} DO}{R} \equal{} 1.$

[b](d)[/b] AC < 2R, BD < 2R. Circumcenter of the $ \triangle ABM$ is $ O \equiv AC \cap BD$ and ABCD is convex $ \Longrightarrow$ $ \angle A < 90^\circ,\ \angle B < 90^\circ \Longrightarrow$ AC > R, BD > R $ \Longrightarrow$ $ AC < 2R < 2BD,$ $ BD < 2R < 2AC.$

[b](e)[/b] $ \frac {S_{ADO}}{S_{ABO}} \plus{} \frac {S_{BCO}}{S_{ABO}} \equal{} \frac {CO}{OB} \plus{} \frac {DO}{OA} \equal{} \frac {CO \plus{} DO}{R} \equal{} 1.$

[b](f)[/b] $ AO \perp BP,\ BO \perp AP$ meet BP, AP at their midpoints A', B'. AP = BP = AB, $ AB' \equal{} PB' \equal{} BA' \equal{} PA' \equal{} \frac {AB}{2}.$ Parallels to AB through C, D meet the normals $ DH_1,\ CH_2$ at $ Q_1, Q_2,$ $ CQ_1 \equal{} DQ_2 \equal{} H_1H_2.$ $ \angle (AP,CQ_1) \equal{} \angle (BP, DQ_2) \equal{} 60^\circ \equal{} \angle CDP,$ $ BC \equal{} CD \equal{} DA \equal{} PC \equal{} PD \Longrightarrow$ the right $ \triangle PB'D \cong \triangle CDQ_1$ are congruent, $ H_1H_1 \equal{} CQ_1 \equal{} PB' \equal{} \frac {AB}{2}.$

[b](g), (h)[/b] See [b](c)[/b].

[u]Proplems is nice and [b]Yetti[/b] is very wonderful. [/u]


Suppose $ M\equal{}\sin10^{\circ}\sin30^{\circ}\sin50^{\circ}\sin70^{\circ}$.

Find $ M$.

:|

[hide="ans"]$ \frac 1 {16}$[/hide]

[hide="Flawed"]
Recall that $ \sin(90 \minus{} x) \equal{} \cos x$. So we have: $ \sin 10\sin 30\cos 10\cos 30 \equal{} (2\sin 10\cos 10)(2\sin 30\cos 30)\cdot\frac {1}{4}$.

$ \sin 20\sin 60\cdot\frac {1}{4} \equal{} \frac {\sqrt {3}}{8}\sin 20$.

$ \sin 20 \equal{} \sin\left(45 \minus{} \frac {30}{2}\right) \equal{} \sin45\cos\frac {30}{2} \minus{} \cos45\sin\frac {30}{2}$.

$ \frac {1}{\sqrt {2}}\left(\sqrt {\frac {1 \plus{} \frac {\sqrt {3}}{2}}{2}} \minus{} \sqrt {\frac {1 \minus{} \frac {\sqrt {3}}{2}}{2}}\right)$.

$ \frac {1}{2\sqrt {2}}\left(\sqrt {2 \plus{} \sqrt {3}} \minus{} \sqrt {2 \minus{} \sqrt {3}}\right)$.

Let $ x \equal{} \sqrt {2 \plus{} \sqrt {3}} \minus{} \sqrt {2 \minus{} \sqrt {3}}$. Then $ x^2 \equal{} 2 \plus{} \sqrt {3} \plus{} 2 \minus{} \sqrt {3} \minus{} 2\sqrt {4 \minus{} 3} \equal{} 2\implies x \equal{} \sqrt {2}$.

So we have $ \frac {\sqrt {3}}{16\sqrt {2}}\cdot\sqrt {2} \equal{} \boxed{\frac {\sqrt {3}}{16}}$.
[/hide]

Where did I go wrong...?

Edit: Hahahahaha, $ 45\minus{}\frac{30}{2}\equal{}30$, not $ 20$.
I'll fix it later.

grn_trtle: Also, sin 70 is not cos 10.
[hide]$ 2M\equal{}\sin 10 \sin 50 \sin 70\equal{}\sin 10 \cos 20 \cos 40$
Multiplying by $ \cos 10$, we have
$ 2M\cos 10\equal{}\frac{1}{2}\sin 20 \cos 20 \cos 40\equal{}\frac{1}{4}\sin 40 \cos 40 \equal{} \frac{1}{8} \sin 80 \equal{} \frac{1}{8} \cos 10$
Therefore, $ 2M\equal{}\frac{1}{8} \iff M\equal{}\frac{1}{16}$[/hide]

[quote="PKMathew"]Suppose $ M \equal{} \sin10^{\circ}\sin30^{\circ}\sin50^{\circ}\sin70^{\circ}$.

Find $ M$.

:|

[hide="ans"]$ \frac 1 {16}$[/hide][/quote]

[hide]
Let $ N\equal{} \cos10^{\circ}\cos30^{\circ}\cos50^{\circ}\cos70^{\circ}$.

Consider $ MN\equal{}\sin10^{\circ} \cos10^{\circ}\sin30^{\circ}\cos30^{\circ}\sin50^{\circ}\cos50^{\circ}\sin70^{\circ}\cos70^{\circ}$.

$ \equal{}\frac{1}{16} \sin20^{\circ}\sin60^{\circ}\sin100^{\circ}\sin140^{\circ}$.

$ \equal{}\frac{1}{16} \sin20^{\circ}\sin60^{\circ}\sin80^{\circ}\sin40^{\circ}$

$ \equal{}\frac{1}{16} \sin20^{\circ}\sin40^{\circ}\sin60^{\circ}\sin80^{\circ}$

$ \equal{}\frac{1}{16} \cos70^{\circ}\cos50^{\circ}\cos30^{\circ}\cos10^{\circ}$

 $ \because \cos x^{\circ}\equal{}\sin(90\minus{}x)^{\circ}$

$ \therefore MN\equal{}\frac{1}{16} N \implies N\left(M\minus{}\frac{1}{16}\right)\equal{}0 \implies M\equal{}\frac{1}{16}$

[/hide]

[quote="Brut3Forc3"]grn_trtle: Also, sin 70 is not cos 10.
[hide]$ 2M \equal{} \sin 10 \sin 50 \sin 70 \equal{} \sin 10 \cos 20 \cos 40$
Multiplying by $ \cos 10$, we have
$ 2M\cos 10 \equal{} \frac {1}{2}\sin 20 \cos 20 \cos 40 \equal{} \frac {1}{4}\sin 40 \cos 40 \equal{} \frac {1}{8} \sin 80 \equal{} \frac {1}{8} \cos 10$
Therefore, $ 2M \equal{} \frac {1}{8} \iff M \equal{} \frac {1}{16}$[/hide][/quote]

That is a very wonderful answer  :) .

[quote="Dr. No"][quote="PKMathew"]Suppose $ M \equal{} \sin10^{\circ}\sin30^{\circ}\sin50^{\circ}\sin70^{\circ}$.

Find $ M$.

:|

[hide="ans"]$ \frac 1 {16}$[/hide][/quote][hide]
\[ M \equal{} \sin10^{\circ}\sin30^{\circ}\sin50^{\circ}\sin70^{\circ} \\
\equal{} \frac12(\cos(30 \minus{} 10) \minus{} \cos(30 \plus{} 10))\cdot\frac12(\cos(70 \minus{} 50) \minus{} \cos(70 \plus{} 50)) \\
\equal{} \frac {1}{4}(\cos 20 \minus{} \cos 40)(\cos 20 \minus{} \cos 120) \\
\equal{} \frac {1}{4}(\cos^2(20) \minus{} \cos 20\cos 120 \minus{} \cos 40\cos 20 \plus{} \cos 40\cos 120) \\
\equal{} \frac {1}{8}((1 \plus{} \cos 40) \minus{} (\cos 100 \plus{} \cos 140) \minus{} (\cos 20 \plus{} \cos 60) \plus{} (\cos 80 \plus{} \cos 160))
\]
[/hide][/quote]

[hide="It may be"]
$ M \equal{} \sin10^{\circ}\sin30^{\circ}\sin50^{\circ}\sin70^{\circ} \\
\equal{} \frac12(\cos(30 \minus{} 10) \minus{} \cos(30 \plus{} 10))\cdot\frac12(\cos(70 \minus{} 50) \minus{} \cos(70 \plus{} 50)) \\
\equal{} \frac {1}{4}(\cos 20 \minus{} \cos 40)(\cos 20 \minus{} \cos 120) \\
\equal{} \frac {1}{4}(\cos^2(20) \minus{} \cos 20\cos 120 \minus{} \cos 40\cos 20 \plus{} \cos 40\cos 120) \\
\boxed{\equal{} \frac {1}{8}((1 \plus{} \cos 40) \minus{} (\cos 100 \plus{} \cos 140) \minus{} (\cos 20 \plus{} \cos 60) \plus{} (\cos 80 \plus{} \cos 160))}$

Here I think it is so cumbersome to simplify.

It is better in this way,
$ \equal{} \frac {1}{4}(\frac{1 \plus{} \cos 40}{2} \plus{} \frac{1}{2}\cos 20 \minus{} \frac{1}{2}(\cos 20 \plus{} \cos 60) \minus{}\frac{1}{2}\cos 40 )$

$ \equal{} \frac {1}{8}(1 \plus{} \cos 40 \plus{} \cos 20 \minus{} (\cos 20 \plus{} \cos 60) \minus{}\cos 40 )$

$ \equal{} \frac {1}{8}(1  \minus{}  \cos 60 )$

$ \boxed{\equal{} \frac {1}{16}}$

[/hide]


man just for debate
who's smarter?
boys or girls.

I pick boys

so boys:1
girls:0

This really is not a necessary topic, and it doesn't go in this IL forum...Plus it can be pretty sexist.

lol. i pick girls because they're hotter than guys  :o

D: girls are known to slow down your mind.

Guys, stop this topic NOW. This is NOT THE MISSOURI FORUM II.

saszs why did you start this thread  :huh:

Saszszzssszssss is like the ultimate spammer.  Seriously, what kind of topic is this, and  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=273619]Saszszsssszszss has a new sibling[/url]??

I disagree, I have seen way more efficent spammers.

Wow,  see the irony in that sentence.

dude does the IL forum even have mods at all? D:

I'm guessing not.. unless they're some kind of hidden mod.  :huh:

Meh state forums get to be cool and less structured :P

yeah, but honestly, this should be about il math, math teams, math competitions in il, ect. Not a free for all wild west where anything goes.

If you can't find a better place to debate the merits of boys versus girls, you honestly haven't tried that hard

[quote="racinginthestreet"]yeah, but honestly, this should be about il math, math teams, math competitions in il, ect. Not a free for all wild west where anything goes.[/quote]

 :clap2:  :clap2:

But in a cool not forcing structure way ;P

Btw are you from IL?

[quote="ra5249"]i think that i should be the il mod[/quote]

Heck,  there is no way that is happening.

I know for a fact that ra isn't from IL. So yeah, he's just a foreign spammer.

*oh gee, I've been gone from AoPS for like a couple of days and I see spam in our lovely little forum* :(  :(  :(

All that sums it up: not good.

That is final.

No more posts in this topic. Please. Ever. Especially posts from spamming foreigners who aren't from Illinois.

True, but my implication is that this is a place to talk about only stuff that relates to Illinoisians. The reason why we are recently annoyed by the non Illinoisians is that they were just spamming the topics (such as saying lies/jokes/sarcasm that isn't really funny; e.g. resurrection and ra saying they go to IMSA, saying they're from IL, wanting to be the IL mod, etc.), which we clearly do not want.

these topics already kinda consist of pointless things. so ya. and i said i failed to get into IMSA =P. the other thing is. every month, chuck norris goes around the world and yeah. I think you get it.

[quote="AIME15"]When you say "your lovely little forum" you imply that the forum belongs to you.  In fact, it is a public forum, and thus anyone can post in it.[/quote]

In fact, it is not a public forum.

[quote="Mewto55555"][quote="gauss1181"]True, but my implication is that this is a place to talk about only stuff that relates to Illinoisians. [/quote]

Illinois borders Missouri, so perhaps spam is a contagious disease?[/quote]

Ooh dang this must have all happened because I just got back from Missouri.

I'M SORRY EVERYONE.

ARML's tonight. If I'm contagious, all of the Midwest will get spammed soon.

nah like indiana is no spam. well. not anymore. OH DANG ITS A DISEASE AND INDIANA GOT IMMUNITY WAY BACK WHEN... that's cool =D

Guys, how about we just let this thread die and not spam anymore in this forum? Thanks.

As was that post.

PLEASE STOP POSTING ON THIS THREAD. IT IS INCREDIBLY IMMATURE AND WILL ONLY WASTE OUR TIME, AS WELL AS DETRACT FROM YOUR REPUTATION AS A VALUABLE CONTRIBUTOR TO THE ILLINOIS FORUM.

THANKS.  :ninja:

[hide]
saszs should either stop spamming or take an aops class if he has so much time.
[/hide]

Yeah that's right if you have enough time on your hands to make topics like this you should go do some more math instead.  It also seems that no matter how hard we try to stop this topic doersn't work,  so I'm not even trying now :P  It'll die eventually.

[quote="Mewto55555"][quote="gauss1181"]True, but my implication is that this is a place to talk about only stuff that relates to Illinoisians. [/quote]

Illinois borders Missouri, so perhaps spam is a contagious disease?[/quote]

I most certainly agree.


I'don't know the category. It even ma be Number Theory...
Given 2003 points inside a cube. Prove that we can divide the cube into more than 2003^3 little cubes such that any of the given 2003 points lies inside one of hte little cubes (not on the faces or sides).

this problem was 1/4 at the Romanian District Olympiad this year (10th grade), but it's not actually a Romanian problem. it's quite simple, so one can say it's a logical problem .... :D

So, if it is easy, what's the solution? :) I think it has to do with Chinese Remainder Theorem, but I may be trying very complicated way of solving it...


geometry
cyclic quadrilateral
analytic geometry
geometry theorems

I think geometry questions are very difficult, because I could hardly come up with any idea to approach the question. Can any experts in geometry give me some advises on approaching a geometry questions?

Thank you very much! :P

I don't feel like being an expert in geometry, but when it comes to solving geometry problems I make a big, neat picture and then gazing at it I try to think of as many theorems I know that might be useful is solving it. Try to toy with it i.e. draw some parallel or perpendicular lines, extend them and so on.
Ask yourself many question like: hmm maybe those three points are collineral or maybe it's a cyclic quadrilateral?. To make the problem look more attractive do not restrict to pencil, but use crayons as well. If you still have no idea how to solve the problem analytic geometry might help very much :lol:


[u][b]On P(Z*) > Z*[/b][/u]

[u][b]General:[/b][/u]

We say that [b]B[/b] > [b]A[/b] if [b]A[/b] not= [b]B[/b] but there is an injection between [b]A[/b] and subset of [b]B[/b] .

Theorem: |[b]P(Z*)[/b]| > |[b]Z*[/b]|

[u]Proof:[/u]

[u]Step 1:[/u] (injection between set [b]A[/b] and subset of set [b]B[/b] ):

Let us define injection between each member of [b]Z*[/b] to each member of [b]P(Z*)[/b], which includes a singleton as its content.

We define f([b]z[/b]) where f([b]z[/b])={[b]z[/b]}, for example: 0 <--> {0} , 1 <--> {1} , 2 <--> {2},… and so on.

At this stage we can say that |[b]P(Z*)[/b]| >= |[b]Z*[/b]| .

[u]Step 2:[/u]

Suppose there exists a bijection between [b]P(Z*)[/b] and [b]Z*[/b] .

We will show that this assumption leads us to contradiction.

There exists member [b]S[/b] in [b]P(Z*)[/b] which is consists of [u][b]ALL[/b][/u] members of [b]Z*[/b] that are not themselves included in the [b]P(Z*)[/b] members which they are mapped to, for example:

0 <--> {0,1} , 1 <--> {10,11,12} , 2 <-->{5,6} , 3 <--> {3,4,5} , 4 <--> {8,9}, …

In this example [b]S[/b] ={1,2,4,…}.

|[b]Z*[/b]|=[i]aleph0[/i]   ,    |[b]P(Z*)[/b]|=2^[i]aleph0[/i]

Bucause {} does not exist in [b]Z*[/b], but {} exists in [b]P(Z*)[/b], then [b]S[/b] has at least one member of [b]Z*[/b], for example:

[b]z[/b] <--> {}

In this example [b]S[/b] ={[b]z[/b]}.

Now, let us say that there exist some member in [b]Z*[/b] (let us call it [b]t[/b] ) which is mapped with [b]S[/b] ([b]t[/b] <--> [b]S[/b] ).

In this case we can ask: is [b]t[/b] in [b]S[/b] or [b]t[/b] not in [b]S[/b] ?

[u]Options:[/u]

1) [b]t[/b] in [b]S[/b] , but by [b]S[/b] definition [b]t[/b] cannot be in [b]S[/b] .

2) [b]t[/b] not in [b]S[/b] , in this case by [b]S[/b] definition, [b]t[/b] must be in [b]S[/b] , but by (1) [b]t[/b] can’t be in [b]S[/b] .

And so on, and so on.

As we can see, both options lead us to logical contradiction.

Therefore, there cannot be a bijection between [b]P(Z*)[/b] and [b]Z*[/b] and we can conclude that |[b]P(Z*)[/b]| > |[b]Z*[/b]| .

Q.E.D

[u][b]A new point of view of this proof:[/b][/u]

By using the empty set (with the Von Neumann Hierarchy), we can construct the set of [b]Z*[/b] numbers {0,1,2,3,...}:

0 = |{ }| (notation = {})

1 = |{{ }}| (notation = {0})

2 = |{{ },{{ }}}| (notation = {0,1})

3 = |{{ },{{ }},{{ },{{ }}}}| (notation = {0,1,2})

...

So, through this point of view the atom is {} and the [b]Z*[/b] numbers are structural|quantitative combinations of {}.

We have here an Hierarchy of infinitely many complexity levels, starting from {}.

If {} does not exist then we have no system to research, therefore we can learn
that in the base of this number system, there is the idea of dependency, which can be translated to the idea of the power of existence.

Let us look at Cantor's proof from this point of view.

As we all know, any set includes only unique members or no members at all.

Now, let us examine options 1 and 2 again.

[u]Option 1:[/u]

All members which included in [b]S[/b] , are different from each other.

Any member of [b]Z*[/b] can be mapped with some member of [b]P(Z*)[/b], once and only once.

Therefore [b]t[/b] is different from each member in [b]S[/b], therefore [b]t[/b]  MUST BE INCLUDED in [b]S[/b].

([b]S[/b] exists iff [b]t[/b] is out of the scope of Cantor's definition. it means that the word [u][b]ALL[/b][/u] is omitted from [b]S[/b] definition).

Therefore [b]S[/b] cannot be defined, and we cannot check our assumption in step 2 of Cantor's proof.

[u]Option 2:[/u]

If we want to keep [b]S[/b] as an existing member, we MUST NOT INCLUDE  [b]t[/b] in [b]S[/b] .

([b]S[/b] exists iff [b]t[/b] is out of the scope of Cantor's definition. it means that the word [u][b]ALL[/b][/u] is omitted from S definition).

In this case we also see that [b]S[/b] cannot be defined, and again we cannot check our assumption in step 2 of Cantor's proof.

[color=blue]Therefore we cannot ask nor answer to anything that is connected to member [b]S[/b], because [b]S[/b]  definition cannot exist, without any connection to the map result between  [b]Z*[/b]  and  [b]P(Z*)[/b].[/color]

The general idea behind this point of view is the power of existence of member [b]t[/b] and member [b]S[/b] .

More you simple less you depended, therefore more exist.

I'll write [b]S[/b] definition again, in an informal way:

There exists member [b]S[/b] in [b]P(Z*)[/b] which is consists of [u][b]ALL[/b][/u] members of [b]Z*[/b] that are not themselves included in the [b]P(Z*)[/b] members which they are mapped to, for example:

0 <--> {0,1} , 1 <--> {10,11,12} , 2 <-->{5,6} , 3 <--> {3,4,5} , 4 <--> {8,9}, …

In this example [b]S[/b] ={1,2,4,…}.

Now, the power of existence of [b]Z*[/b] members is bigger than the power of existence of [b]P(Z*)[/b] members.

Member [b]t[/b], which is some arbitrary member of [b]Z*[/b] (and mapped with [b]S[/b]), is simpler than [b]S[/b], which is a member of [b]P(Z*)[/b] .

Member [b]S[/b] existence depends on objects like [b]t[/b] , therefore we have to check [b]S[/b] by [b]t[/b] as we did here, and not [b]t[/b] by [b]S[/b] as Cantor did.

[u]Conclusion: [/u]Cantor did not prove that |[b]P(Z*)[/b]| > |[b]Z*[/b]| because the definition  [b]S[/b]={[b]z[/b] in [b]Z*[/b] such that [b]z[/b] is not in f([b]z[/b])} cannot exist because  [b]P(Z*)[/b] cannot exist without [b]Z*[/b].

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

{{}} can exist only if {} is nested in another {}.

In other words, there is a hierarchy of dependency of composed sets in non-composed sets, and this is exactly the deep meaning of the word ‘hierarchy’ in Von Neumann Hierarchy, which its unique nested structure is based on 3 [b]ZF[/b] axioms:

[b]ZF[/b] pair axiom, [b]ZF[/b] sum axiom and [b]ZF[/b] infinity axiom.

Pair and sum axioms are its recursive side and infinity axiom is its inductive side.

In other words, there is no “deep” thing here but some arbitrary way to construct nested sets, by using the empty set as their building block, for example:

If → means ‘implies’ then {} → {{}} → {{},{{}}} → {{},{{}},{{},{{}}}} etc. where the existence of the more composed element depends on the existence of the less composed element.

Also a non-nested collection is simpler than a nested collection, even if they have the same cardinal, for example:

|{{}}| is the same as |{{}}|
|{{},{}}| is simpler than |{{},{{}}}|
|{{},{},{}}| is simpler than |{{},{{}},{{},{{}}}}|
|{{},{},{},{}}| is simpler than |{{},{{}},{{},{{}}},{{},{{}},{{},{{}}}}}|

… etc.

Simplicity is one of the most powerful attitudes that can be used in order to understand fine concepts, and I used simplicity in order to research non-empty finite and non-finite collections.

The simplest way to define the unique cardinal of each [b]n[/b] in [b]N[/b], is based on the quantity of the empty sets that can be found within each [b]n[/b].

For example:

[code]
              {}

           {} {}

        {} {} {}

     {} {} {} {}

{ {},{},{},{},{}, … }

   1  2  3  4  5

   ↕  ↕  ↕  ↕  ↕

   2  4  6  8 10

{ {},{},{},{},{}, … }

  {} {} {} {} {}

     {} {} {} {}

     {} {} {} {}

        {} {} {}

        {} {} {}

           {} {}

           {} {}

              {}

              {}

[/code]
If we examine the 1-1 mapping between each [b]N[/b] member and each member of a proper subset of [b]N[/b] (notated as [b]S[/b]) of even members, we realize that set [b]S[/b] is nothing but a particular way to pack the empty sets, which is based on a successor of exactly two empty sets.

Let a successor be the result of card(nth [b]n[/b])-card(nth-1 [b]n[/b]) where (nth-1 [b]n[/b])<(nth [b]n[/b]).

If nth-1=0 then [b]n[/b] has no successor.

Actually we can define ordered non-finite sets of [b]N[/b] members, which are different from each other by the sequences of their successors.

But we have to understand that no matter what combination we use, it does not change the 1-1 mapping between a collection of infinitely many empty sets, into itself, for example:

[code]
{ {},{},{},{},{},{},…  }

  ↕  ↕  ↕  ↕  ↕  ↕

{ {},{},{},{},{},{},…  }
[/code]
If we take the above example of vertical collections of [b]N[/b] members, which are constructed by empty sets, and order them in a  horizontal way, we clearly realize that [b]Peano[/b] or [b]ZF[/b] axioms ignore the empty sets themselves and use a 1-1 mapping between [b]N[/b] members, which are actually constructed by packs of empty sets.

The Cantorean 1-1 mapping ignores the level of empty sets, for example:

[code]
{  {},    {{}{}},     {{}{}{}},       {{}{}{}{}}, ... }

   ↕       ↕           ↕               ↕

{ {{}{}}, {{}{}{}{}}, {{}{}{}{}{}{}}, {{}{}{}{}{}{}{}{}}, ... }
[/code]
By ignoring the most fundamental level of the empty sets, we can get the illusion that {2,4,6,8,...} is a proper subset of {1,2,3,4,...}.

When we ignore the packed level of both {2,4,6,8,...} and {1,2,3,4,...} and define 1-1 mapping between the empty sets that are included within each natural number, we realize that we actually deal with the same non-finite collection of empty sets.

Furthermore, from this basic level, we also realize that a non-finite collection has no identity map because its 1-1 mapping cannot be satisfied.

In order to understand why, please look at:

http://www.geocities.com/complementarytheory/Successor.pdf

Also please look at:

http://www.createforum.com/phpbb/viewtopic.php?t=20&mforum=geproject

[quote="jmerry"]This appears to be the work of a crank.[/quote]

He's got posts on at least a half-dozen forums for math, science and philosophy on his theories.  The point, apparently, is that he is creating a new reality of some sort:

[quote="DoronShadmi from another forum"]What I show is that there is a deeper point of view of the same elements, where the "old" proofs become trivial (trivial not in the mathematical meaning) because they ignore the internal complexity of the examined elements...
I do not  think in terms of 'right'_XOR_'wrong', but in terms of the triviality and non-interesting results of these proofs, when they are understood from a non-trivial point of view.[/quote]


A sequence $a_{n}$ is defined as follows, $a_{0}= 1$, $a_{n+1}= (1+4a_{n}+\sqrt{(1+24a_{n})})/16$ for $n \ge 0$. Find an explicit formula for $a_{n}$.

[quote="shobber"]You mean $4a_{n}$ or $4a \cdot n$?[/quote]
It is certainly $4a_{n}$ as in this case $a_{1},a_{2},a_{3}...$ come out quite nicely. In addition, $4a \cdot n$ doesn't make much sense in the problem.

Today when I was reading a book, I accidently found out that this problem is from the Contest of the West Germany of year 1986.


analytic geometry
graphing lines
slope
calculus
Ross Mathematics Program
MIT
college

What exactly is an example of rigor in a proof being carried too far?

I'm thinking that the most rigorous proof in say, real analysis, would also require proofs in number theory - a proof that would have to specify why 1 + 1 = 2, 5 + 5 = 10, and so on. The analysis of a certain type of graph would have to specify why the graph is the way it is, why the graph f(x) is always continuous, why it always has a constant slope, why the mapping is 1-1.

The other question is...what exactly is rigor? I mean - the axioms don't have to be rigorous themselves. They must only be consistent - but you can just specify more axioms that have been proven as axioms. Couldn't the axioms of calculus been axioms in the first place - and then be tackled down and reduced to even more fundamental axioms? How fundamental can it get?

I mean, when Newton and Leibnitz defined calculus, how could they be sure that calculus wasn't completely rigorous? It lies in the definitions. So then, the properties of number theory lie in some definitions - and the properties that we see in a textbook may still not be completely rigorous.

Does anyone know the content of the Ross summer program problem sets? I remember seeing them a few years ago - but now I can't google any of them. I heard that they spent an immense amount of time proving some of the most basic properties of number theory. Such as... why a + 0 = a. or something like that.

this question is purely theoretical and out of interest. I'm curious to know how rigorous is "too rigorous?"

[quote]What exactly is an example of rigor in a proof being carried too far? [/quote] You can find one in [i]A Mathematician's Miscellany[/i] by Littlewood (a proof of the Weierstrass approximation theorem).

[quote]a proof that would have to specify why 1 + 1 = 2, 5 + 5 = 10[/quote] And before "why?" comes the question "what?": what is $1$, what is $5$, etc. You may want to read [url=http://en.wikipedia.org/wiki/Principia_Mathematica]Principia Mathematica[/url]. (Just kidding!  :rotfl: )

I'll offer my biased advice: don't bother looking for "a solid foundation of mathematics" in a system of axioms. It isn't there.

hmm, true.

Also, say...

[url=http://ocw.mit.edu/NR/rdonlyres/Mathematics/18-014Calculus-with-Theory-IFall2002/83860E28-54FD-4CF5-A8E4-9793DB3831EF/0/pset3solutions.pdf]This MIT problem set solution[/url]

This is just a sample. Anyways, it's based on proofs. But it's not a completely rigorous proof. The fact is - it would have to prove the Fundamental Theorem of Calculus in order to be completely rigorous. It would have to argue that this problem is not different from a problem in integration. It would also have to specify how addition can be proved, how subtraction of functions can be proved, how the additivity of functions can be proved.

Have we even proved say, the Fundamental Theorem of calculus to the maximum depth possible? We needed mathematicians after Newton to prove that - it needs epsilon delta definitions. And are epsilon delta definitions used in number theory?

The question is - when are people usually satisfied with the rigor of a proof? The amount of rigor is unspecified, and it seems that people's intuitions are usually good enough to appreciate just how rigorous a proof would be.

[i]Edited by moderator: please don't post very long url-s.[/i]

I may have more on this later, but you seem to be working under the principle that referencing theorems is not rigorous.  This is absurd: if theorem $X$ has a rigorous proof, there is no reason I should repeat the proof every time I use the theorem.  A result does not become less true if I fail to provide its proof.

People often don't reference theorems in such proofs though. (especially the ones provided in the problem set solutions)

A proof like this ( http://mathforum.org/library/drmath/view/51551.html ) certainly isn't referenced in such problems. And a fully rigorous proof would have to be able to reference this first (well, being pedantic).

By the way, how rigorous can number theory ultimately become?

I think rigor is what is accepted by the majority of sane experts in the field. If all (or most) mathematicians accepted some seemingly absurd result, then it would have to be accepted. Now you may ask, what is sane? Well I think that's for you to decide, because a mentally insane person probably believes that they are the only rational mind.
Rigor can also be defined as "what is agreed upon". Upon questioning of your proof, the reader himself should be able to completely answer all the questions would arise, given that he has a complete knowledge of previous work that you are citing. Rigor is lack of ambiguity, and level of rigor can probably be correlated to "amount of questioning" that the reader would have. If after going over your paper, a reader completely understands your assertions, and has no questions about the contents themselves (as opposed to say, proposing extensions), then there is no reason to dock any "rigor points".

This might be a bit off topic, but I believe someone once said that mathematicians do work to challenge and interest fellow mathematicians only. I think this relates somewhat to this discussion :D

To be rigorous, one has just to name the theorem one uses, not to give a reference everytime. Thus writting $1+1=2$ only already contains that we already know this to be true.
A proof is rigorous if we know all used facts to be already true. But it may sometimes be helpfull to give a reference.

And Number Theory is not less or more rigorous than any other part of mathematics (thus totally rigorous).
By the way, $1+1=2$ is not number theory (you speak like number theory is just how to handle integers etc.).

I can give several examples of how rigorous is "too rigorous".  Simply open any middle/high school geometry textbook and look for the two-column proofs.  Surely everyone remembers these--the ones where you have to cite the reason for every step you take next to the step (and if there is more than one reason, it must be more than one step :roll: ).  I used to hate those, mainly because they are the epitome of rigor carried to unnecessary extremes.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9991]An older thread about the nature of proof[/url]


geometry
inradius
inequalities
circumcircle
trigonometry
geometry solved

Let $r$ be the inradius of an acute triangle. Prove that the sum of the distances from the orthocenter to the sides of the triangle does not exceed $3r$.

Erdos-Mordell's inequality gives the weaker bound $\leq\frac{3}{2}\cdot R$ instead of $\leq 3r$. Hence, applying Erdos-Mordell here is not good enough. Now the problem becomes very interresting.

Now here is my solution
We will firs prove that if $a\leq b\leq c$ then $d_a\leq d_b\leq d_c$ where $d_a$ is
distance from orthocenter to side $a=BC$ and so on. If $a\leq b$ then $\angle A\leq
\angle B$ how $\angle B'AA'=\angle B'BA'$ ($A'$ is foot of altitude from $A$ and so
on) then $\angle HAB\leq \angle HBA$ now $\angle HAB=\angle HB'A'$ and $\angle
HBA=\angle HA'B'$ you obtain $d_a\leq d_b$. Now we use that inequality i don't know
to write name but i'll try "Chebyshev", (hope that you know what I mean)

Because $a\leq b \leq c$ implies $d_a\leq d_b \leq d_c$ we use
\[3(ad_a+bd_b+cd_c)\geq(a+b+c)(d_a+d_b+d_c)\]

left side is equal $6P$ where $P$ is area of triangle so we have \[6P\geq
(a+b+c)(d_a+d_b+d_c)\] and from that, because $P=(a+b+c)\frac{r}2$ we obtain
\[3r\geq (d_a+d_b+d_c)\] $Q.E.D$

I will be glad to hear others

m@re :) :D

M@re's solution is really beautiful; as he asked for different ones, I am posting mine. It is way longer, but yields slightly more.

[color=blue][b]Problem.[/b] Let ABC be an acute-angled triangle, and let its altitudes from the vertices A, B, C have the feet A', B', C', respectively, and intersect at the orthocenter H of triangle ABC. Let r be the inradius and R the circumradius of triangle ABC. Then,

[/color][color=green][b](1)[/b][/color][color=blue] $HA^{\prime}+HB^{\prime}+HC^{\prime}\leq 3r$.[/color]

[i]Solution.[/i] According to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24870 Lemma 1, we have $AH=2R\cos A$. Now, in the right-angled triangle AA'C, we have < A'AC = 90° - < ACA', what rewrites as < HAB' = 90° - C; thus, in the right-angled triangle AB'H, we have $HB^{\prime}=AH\cdot\sin\measuredangle HAB^{\prime}=2R\cos A\cdot\sin\left(90^{\circ}-C\right)=2R\cos A\cdot\cos C$. So we have obtained $HB^{\prime}=2R\cos C\cos A$. Similarly, $HC^{\prime}=2R\cos A\cos B$ and $HA^{\prime}=2R\cos B\cos C$. Thus, the inequality [color=green][b](1)[/b][/color] becomes $2R\cos B\cos C+2R\cos C\cos A+2R\cos A\cos B\leq 3r$, or, equivalently, $\cos B\cos C+\cos C\cos A+\cos A\cos B\leq\frac{3r}{2R}$. Hence, the problem will be solved once the following theorem will be verified:

[color=blue][b]Theorem 1.[/b] If ABC is an arbitrary triangle with inradius r and circumradius R, then $\cos B\cos C+\cos C\cos A+\cos A\cos B\leq\frac{3r}{2R}$.[/color]

Note that in Theorem 1, triangle ABC needs not to be acute-angled anymore like in the problem; this is why I claimed that I will prove slightly more than the actual problem.

[i]Proof of Theorem 1.[/i] Using the notation $\sum$ for cyclic sums, the inequality from Theorem 1 becomes

[color=green][b](2)[/b][/color] $\sum\cos B\cos C\leq\frac{3r}{2R}$.

Now, according to the well-known formula $\frac{r}{R}=\cos A+\cos B+\cos C-1$ (proven in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=290190#p290190]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45622 from post #5 on[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=370715#p370715]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61455[/url]), we have

$\frac{3r}{2R}=\frac32\cdot\frac{r}{R}=\frac32\cdot\left(\cos A+\cos B+\cos C-1\right)=\frac32\left(\sum\cos A-1\right)$
$=\frac32\left(\sum\left(1-2\sin^2\frac{A}{2}\right)-1\right)=\frac32\left(\left(3-2\sum\sin^2\frac{A}{2}\right)-1\right)$
$=\frac32\left(2-2\sum\sin^2\frac{A}{2}\right)=3-3\sum\sin^2\frac{A}{2}$,

while

$\sum\cos B\cos C=\sum\left(1-2\sin^2\frac{B}{2}\right)\left(1-2\sin^2\frac{C}{2}\right)$
$=\sum\left(1-2\sin^2\frac{B}{2}-2\sin^2\frac{C}{2}+4\sin^2\frac{B}{2}\sin^2\frac{C}{2}\right)$
$=3-4\sum\sin^2\frac{A}{2}+4\sum\sin^2\frac{B}{2}\sin^2\frac{C}{2}$.

Thus, the inequality in question, [color=green][b](2)[/b][/color], becomes

$3-4\sum\sin^2\frac{A}{2}+4\sum\sin^2\frac{B}{2}\sin^2\frac{C}{2}\leq 3-3\sum\sin^2\frac{A}{2}$,

or, equivalently, $\sum\sin^2\frac{A}{2}\geq 4\sum\sin^2\frac{B}{2}\sin^2\frac{C}{2}$. Division by $\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$ transforms this into

[color=green][b](3)[/b][/color] $\sum\frac{\sin\frac{A}{2}}{\sin\frac{B}{2}\sin\frac{C}{2}}\geq 4\sum\frac{\sin\frac{B}{2}\sin\frac{C}{2}}{\sin\frac{A}{2}}$.

Now, if $s =\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC, then, by the half-angle formulas,

$\frac{\sin\frac{A}{2}}{\sin\frac{B}{2}\sin\frac{C}{2}}=\frac{\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{bc}}}{\sqrt{\frac{\left(s-c\right)\left(s-a\right)}{ca}}\cdot\sqrt{\frac{\left(s-a\right)\left(s-b\right)}{ab}}}=\frac{a}{s-a}$;

by taking the reciprocal, we get $\frac{\sin\frac{B}{2}\sin\frac{C}{2}}{\sin\frac{A}{2}}=\frac{s-a}{a}$. Thus, the inequality [color=green][b](3)[/b][/color] becomes

[color=green][b](4)[/b][/color] $\sum\frac{a}{s-a}\geq 4\sum\frac{s-a}{a}$.

But this is pretty easy to prove: Since (s - b) + (s - c) = 2s - (b + c) = (a + b + c) - (b + c) = a, we have

$\sum\frac{a}{s-a}=\sum\frac{\left(s-b\right)+\left(s-c\right)}{s-a}=\sum\frac{s-b}{s-a}+\sum\frac{s-c}{s-a}$
$=\sum\frac{s-a}{s-c}+\sum\frac{s-a}{s-b}=\sum\left(s-a\right)\left(\frac{1}{s-b}+\frac{1}{s-c}\right)$.

Since $\frac{1}{s-b}+\frac{1}{s-c}\geq\frac{4}{\left(s-b\right)+\left(s-c\right)}$ by AM-HM or Cauchy-Schwarz, we thus have

$\sum\frac{a}{s-a}=\sum\left(s-a\right)\left(\frac{1}{s-b}+\frac{1}{s-c}\right)\geq\sum\left(s-a\right)\frac{4}{\left(s-b\right)+\left(s-c\right)}$
$=\sum\left(s-a\right)\frac{4}{a}=4\sum\frac{s-a}{a}$.

Hence, the inequality [color=green][b](4)[/b][/color], and with it Theorem 1 is proven.

  Darij


A palindrome is a number that reads the same forward and backward, such as 212. What is the probability that a three digit positive integer is a palindrome? Express your answer as a common fraction.

The number must be in form of $ aba$.  The first digit and second digit both don't matter.  However, the third digit must be the same as the first, so the answer is $ \boxed{\frac{1}{10}}$.

Or we could list:

101
111
121
131
141
151
161
171
181
191
etc. We see that there are 10 per hundred,so 1/10

I like Ernie's solution more, but you could also think of it like this: 9 choices for the first digit (can't use 0), 10 for the second digit (digits can be reused), and third is locked in because it must be the same as the first. $9*10=90$ unique palindromes, and 900 three digit numbers. So, we get $90/900 = 1/10$


A simple way to get pi is $ \pi \equal{} \frac {4}{1} \minus{} \frac {4}{3} \plus{} \frac{4}{5} \minus{} \frac {4}{7} \plus{} \frac {4}{9}...$. As you can see, there's a pattern. If you keep on doing this formula, you'd eventally start seeing digits of pi.

 So I'll first start with $ \frac {4}{1} \minus{} \frac {4}{3}$. You can see that the denominators are odd numbers in sequence, so the next person would ADD $ \frac {4}{next \ odd \ number}$. Then the person after that would SUBTRACT $ \frac {4}{next \ odd \ number}$. If you don't understand how to do this, just tell me and I'll explain it to you.

$ \frac {4}{1} \minus{} \frac {4}{3} \equal{} \frac {9}{3} \equal{} 3.$ Then do the above, turn the fraction to make it a decimal, then add it to the total and display it like this.

Current Total: $ \frac{8}{3}$

[quote="Cal Culator"]A simple way to get pi is $ \pi \equal{} \frac {4}{1} \minus{} \frac {4}{3} \plus{} \frac {4}{7} \minus{} \frac {4}{9}...$. As you can see, there's a pattern. If you keep on doing this formula, you'd eventally start seeing digits of pi.

 So I'll first start with $ \frac {4}{1} \minus{} \frac {4}{3}$. You can see that the denominators are odd numbers in sequence, so the next person would ADD $ \frac {4}{next \ odd \ number}$. Then the person after that would SUBTRACT $ \frac {4}{next \ odd \ number}$. If you don't understand how to do this, just tell me and I'll explain it to you.

$ \frac {4}{1} \minus{} \frac {4}{3} \equal{} \frac {9}{3} \equal{} 3.$ Then do the above, turn the fraction to make it a decimal, then add it to the total and display it like this.

Current Total: 3[/quote]

Erm...

$ \frac {4}{1} \minus{} \frac {4}{3} \equal{} \frac {8}{3}$

This sequence converges extremely extremely slowly to pi. You might want to put decimal approximations at each partial sum to actually see how this series converges...

Also, this should definitely be moved to Fun Factory. Not that I would call this any particular sort of fun...  :huh:

Hmm... I found out that doing this way calculates pi very, very slow. It would take about 300 sequences to calculate correctly 3 digits of pi (3.14. This is according to wikipedia). There is a slightly more complex, but faster way to do it based off this series (you can use it to prctice math skills :lol: ). It goes like this:

Let $ \pi_{0,1} \equal{} \frac{4}{1}, \pi_{0,2} \equal{} \frac{4}{3}$ etc.
and then define $ \pi_{i,j}\equal{}\frac{\pi_{i\minus{}1,j}\plus{}\pi_{i\minus{}1,j\plus{}1}}{2}$ and for all  $ i,j$ greater then or equal to 1. You should get pi correct to nine digits if you calculate $ \pi_{10,10}$.
I don't understand how to do this though.

Luckily I Don't know how to delete this topic  :) . I'm woking on how to find i, because it starts at 0. If anyone knows how to do it, please let me know how.

Because the first term is appreciably larger than pi (4=4.0000....) and the second term is appreciably smaller (8/3=2.6666...), why not average them (arithmetic mean). You get the result, (3.333333....) which is clearly a better estimate.

This is (1,1). (1,2) turns out to be quite a bit smaller than pi still: (3.0666...)
We can calculate (2,1) by averaging (1,1) and (1,2) - giving us (3.2), it is clearly a much better estimate than the previous values.

The reason this works is that the estimate fluctuates, first above pi, then below pi. If the original sequence converges to pi, then subsequent averaging result in faster approximations.


find all $x,y,z$ is integer number and $p$ is prime such that  $x^y=p^z+1$


function
algebra
domain
complex analysis

I have a quick question that I suspect is true but I am not positive.

Suppose $ f$ and $ g$ are two meromorphic functions. We know that $ z_0$ is a pole for $ f$ iff it is a pole for $ g$ and the order of the pole for $ f$ at $ z_0$ is the same as the order of the pole at $ z_0$ for $ g$ (although we don't know that they necessarily share the same coefficients $ a_{\minus{}n}$ in their Laurent expansion at $ z_0$). Furthermore, $ f$ and $ g$ agree on the reals.
Does it follow that $ f \equal{} g$?

Thanks for the help

You don't need to know anything about the pole to say that- agreement on a set with a non-singularity accumulation point is enough (assuming the intersection of domains is connected). There is no practical difference between meromorphic and analytic functions here, because meromorphic functions are analytic except at isolated points.


What do we need to consider when we are going to apply AM-GM?
Just like
Determine the minimum value of f(a) when a is [b]non-negative[/b]
$f(a)=a^{2}+a+b$
where b is a [b]constant[/b].
If we apply the first one, we have the minimum value when
$a^{2}=a$
then, if we are still couldn't see the solution, we may go to the next step,
so we got $a^{(3/2)}=b$
that mean b can only be 1 or 0.
Hence, the solution is not correct.
Sometimes, when it involves a lot of variables, a b c d;
here, Again the same question
What do we need to consider when we are going to apply AM-GM?

Thanks

I can't understand what you mean. Can you explain more clearly?

OK. Appreciate for your quick response.
When I am using AM-GM solving inequalities, I found that we can't use it infinitely. As illustrated in the example above, by the first AM-GM, we got a being 0 or 1.
In fact, f(a) reaches its minimum value when a is 0.
But, if we don't stop here, thinking that we can use another AM-GM once.
Then, we got the wrong solution.
So my question is:
What besides non-negative numbers do we consider when we are applying the AM-GM inequalities?

You should consider when the equality condition holds. I mean, if you used some inequalities to prove inequality or find minimum/maximum, the equality condition of whole inequality is the intersection of all of the equality conditions of the partial inequalities.

For example, you can think that the minimum of $(a+\frac{1}{b})(b+\frac{4}{a})$ is 8, when $a,b>0$, as $a+\frac{1}{b}\geq 2\sqrt{\frac{a}{b}}$ and $b+\frac{4}{a}\geq 4\sqrt{\frac{b}{a}}$. However, there is no equality condition. Actually this expression has minimum 9 as $(a+\frac{1}{b})(b+\frac{4}{a})=5+ab+\frac{4}{ab}\geq 5+4=9$.

Moreover, in case of youe inequality, when you tried to find a minimum of the expression, you sholud consider the proper AM GM, not any form, so that the result does not involve any variables (such as $a$) and there should be equality case which can satisfy all equalities of the partial inequalities.

Your example illustrates my mind a hundred times better than mine.
Thank you very much.


combinatorics unsolved
combinatorics
ez
easy

In a circus, there are $n$ clowns who dress and paint themselves up using a selection of 12 distinct colours. Each clown is required to use at least five different colours. One day, the ringmaster of the circus orders that no two clowns have exactly the same set of colours and no more than 20 clowns may use any one particular colour. Find the largest number $n$ of clowns so as to make the ringmaster's order possible.

trivial stuff...20*12/5=48...the example is trivial as well...(fix 4 at a time)

Perhaps the most bizarre thing is that according to the official solution, the bound is worth 5 (!) marks, and the construction is worth 2.

What is the construction for $48$ colors? What is the motivation for finding it?

Complete solution (with construction) of this problem is published here on page 48.

[url]http://estoyanov.net/files/MATAMATIKA/aziatsko_tihookeanski_matholimpiadi.pdf[/url]

But I would also like to know the motivation.

Simplest solution....
Let $a_1$,$a_2$,.....$a_n$ denote the $n$ clowns and let $c_1$,$c_2$,.....,$c_{12}$ denote the 12 colors.

Set up the incidence matrix. An entry is 1 if the clown corresponding to a color uses that color, and 0 otherwise. Count the no. of ones in two different ways. If N is the no. of ones, then we get $5n\le N\le 240$. From this inequality, we find our desired answer to be $48$.

Here is easy example take $6$ groups of $8$ clowns .Mark these groups from $1$ to $6$ .In first group all clowns have form  $1,2,3,4,k$  where $k$ is one of other 8 not used ($5\cdots,12$ in this case) (k will be defined similarly for others  ) .Every clown in $2$ has form $5,6,7,8,k$.Third has $9,10,11,12,k$.Fourth $1,2,5,9,k$ ; Fifth $3,6,7,10,k$ and last one $4,8,11,12,k$


Given that $ a,b,c$ are positive reals, prove that:
a)
\[ \frac {2(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc)}{a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b)} \minus{} \frac {1}{2} \geq \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}
\]
b)
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \geq \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}{a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b)} \plus{} \frac {1}{2}
\]

[quote="shaaam"]Given that $ a,b,c$ are positive reals, prove that:
b)
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \geq \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}{a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b)} \plus{} \frac {1}{2}
\]
[/quote]

We just need to rewrite the inequality in the true form. :)

$ \sum_{cyc} {\frac {{a}}{{b \plus{} c}}} \minus{} \frac{3}{2} \ge \frac {{a^3 \plus{} b^3 \plus{} c^3\plus{}3abc }}{{\sum_{sym}{a^2b}}} \minus{} 1$
or

$ \frac {2{\sum_{cyc} {a^3 }\minus{} \sum_{sym} {a^2b } }}{{2\sum_{sym}{a^2b} \plus{} 4abc}} \ge
\frac {{\sum_{cyc} {a^3 } \plus{}3abc\minus{} \sum_{sym} {a^2b} }}{{\sum_{sym} {a^2b}}}$
Let's get:
$ x\equal{}\sum_{cyc} {a^3}$
$ y\equal{}\sum_{sym}{a^2b}$
$ z\equal{}abc$
then we have:
$ \frac{2x\minus{}y}{2y\plus{}4z} \ge \frac{2x\plus{}3z\minus{}y}{y}$
or
$ y^2 \ge 4xz\plus{}12z^2\plus{}2yz$
or
$ \sum_{sym}{x^4y^2}\plus{}\sum_{sym}{x^3y^3}\plus{}\sum_{sym}{x^4yz}\plus{}\sum_{sym}{x^2y^2z^2}\plus{}2\sum_{sym}{x^3y^2z} \ge 2\sum_{sym}{x^4yz}\plus{}2\sum_{sym}{x^2y^2z^2}\plus{}2\sum_{sym}{x^3y^2z}$
$ \Leftrightarrow$
$ \sum_{sym}{x^4y^2}\plus{}\sum_{sym}{x^3y^3} \ge \sum_{sym}{x^4yz}\plus{}\sum_{sym}{x^2y^2z^2}$

which is obvious. :)


Prove the line joining the midpoints of 2 sides of a triangle is parallel to the third side.

[hide="HINT"]
Similar triangles.
[/hide]

[quote="AstroPhys"]
Similar triangles.
[/quote]

Do you use the SAS rule to prove that they are similar


Let $ S$ be the set of points $ (a,b)$ in the coordinate plane, where each of
$ a$ and $ b$ may be $ \minus{}1$, 0, or 1. How many distinct lines pass through
at least two members of $ S$?

Anybody plz post a simple solution that I can easily understand.

There are $3$ vertical lines, with slope undefined.
There are $3$ horizontal lines, with slope $0$.
There are $3$ diagonal lines with slope $1$.
There are $3$ diagonal lines with slope $-1$.

So in total, there are $\boxed{12}$ lines that pass through at least $2$ members of $S$.

[hide=Solution]
With $9$ points, there are $\binom{9}{2}=36$ ways to choose $2$ of them, which would determine a line. However, for any three collinear points, we are over counting, and must subtract by twice the number of cases in which we overcount because there are three collinear points, $\binom{3}{2}=3$ ways to choose two of them, and only one time we want to count the line. We are over counting the vertical lines $x=\pm 1,0$, the horizontal lines $y=\pm 1, 0$, and the lines $y=\pm x$, because they cross three of the points. This is $8$ lines, so we need to multiply this by two to get $16$, and subtract this from $36$, which brings us to our final answer of $\boxed{20}$. [/hide]

[hide=Solution]We know that $S$ is a 3x3 grid of dots:
[asy]
size(1cm);
pair A = (1,0);
pair B = (0,0);
pair C = (-1,0);
pair D = (-1,-1);
pair E = (-1, 1);
pair F = (1,-1);
pair G = (1,1);
pair H = (0,1);
pair I = (0,-1);

dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(F);
dot(G);
dot(H);
dot(I);
[/asy]
To find the total number of pairs of points, we have $\binom{9}{2}$. However, we overcount the lines that pass through more than two members of $S$. There are 3 horizontal lines and 3 vertical lines that pass through three points, as well as 2 diagonal lines. So, since we count each line with three points $\binom{3}{2}$ times, we subtract $8$ twice from $\binom{9}{2}$. Our answer is $$\binom{9}{2}-8(2)=\boxed{20}.$$
[/hide]


It took them longer than usual this year but now they totally changed the interface of the website. I don't know how long they had the video and the new interface up for but heres the video
http://216.235.203.102/NETCOMMUNITY/Page.aspx?pid=535&srcid=316

Just a notice that you can clearly see our whole team (Carl, me, Elizabeth, David) when we're going to the written competition room. Also, there was a kid in pink in the background near the beginning which was probably part of the MA team also... You also see Elizabeth giving a small speech at the end and in the countdown but I think that's it. You see a bunch of other people who is an AOPSer also. Overall, the video is quite a bit longer than usual and I find the announcement of the top 3 teams to be a little anticlimactic.

In other news, check out mathcounts.org and they have a BUNCH of new stuff coming, like a live broadcast of the 2008 Nats. It's going to be pretty interesting. I'll probably take a look at the webcast.

At chapter I had my team cheer when they showed us in our video last year...

Kevin looked a bit weird on the last CD question.
It looked like he had the answer, but wasn't sure whether he wanted to buzz in or not...

Omg, they are too fast...
But then I wasn't able to see the questions...

yeah dude the kid in the pink shirt is definitely me.

ehhhhh at the end, if they show the top 10 teams, they should definitely show the top 40 individuals. it makes the ratios consistent, for number announced before: number announced after. or whatever, as long as it's more than 15.

gah seeing this reminds me of how much i screwed myself over that day and also how much fun i had. every time i hear country music (like today, for example), i get reminded of that trip. like no joke. because of the stockyards (which by far was the worst part of the trip) and the continuous country music at the hotel.

Does it remind you of getting lost iin the Maze?  Standing on the watch tower with david and watching you try to find your way out was my favorite part :lol:   The rodeo was the worst part...well that singer was the worst part...or waiting in line for that night's dinner.

[quote="CatalystOfNostalgia"]The maze was aiight, waiting for food was relatively worth it, while the country singer has dotted my nightmares for the past year or so.

What is the Friday night trip in Denver? Skiing?[/quote]

lol, no.  We are having dinner in a museum.  I think they did the same thing your 7th grade year in the aeronautical museum.  Dinner in a museum is often the Friday night plan.

The best part is probably the maze or the place with the candy shop. I still remember Carl riding some plastic pony and when he spent (wasted) like a dollar on this fortuneteller machine.

[quote="perfect628"][quote="frost13"], no.  We are having dinner in a museum.  I think they did the same thing your 7th grade year in the aeronautical museum.  Dinner in a museum is often the Friday night plan.[/quote]
Yep--we played ghost and a "game" where you had to keep going around naming different country capitals. Oh and we raced the elavator vs. running on the stairs.

[quote="Moldytape"]I still remember Carl riding some plastic pony [/quote]
WHAT[/quote]
Hmm, I can't imagine Carl riding a pony...

Me either.

The movie was cool though, except it should have had more countdown (at least a better view).

Techincally the stockyards was dinner in a museum. Museum of stupidity, horses, and people with no lives.

Rodeo = 3D movie on Mars rovers. Both pinnacles of modern intelligence.

I seem to remember the plastic pony as well. The fortune teller, well, I remember it vaguely, you had to remind me. Dang, that Saturday at the Stockyards sucked...you guys basically spent an hour in the candy shop after we had gone THE NIGHT BEFORE for some reason, while I sat on the bench outside and relived my suxx0r4g3 from the day before. Then we ate in that place where I spent 5 bucks on a plate of cheese fries that was unpredictably big (seriously, wtf type of appetizer was that), and ate like 1/3 of it. I did eat my whole burger though. And hey that came with fries, I believe.

But it got better because after we attempted playing frisbee on tennis court in 105 degree weather, we got all of our stress out in the pool.

Dude Sam, I believe you don't remember the game correctly, the game was to ask "Are we in [capital]?" until we either ran out of ideas so resorted to Washington D.C., in which the answer was yes, or got really bored. Sadly, the first happened.

Yeah, the stairs won every time. It wasn't even a contest. Dude, I burned so many calories that night. But the 3D movie was good. Arlington National Cemetary was good. The Spy Museum was aiight. Cingular signs = God.

[quote="CatalystOfNostalgia"]Dude Sam, I believe you don't remember the game correctly, the game was to ask "Are we in [capital]?" until we either ran out of ideas so resorted to Washington D.C., in which the answer was yes, or got really bored. Sadly, the first happened.[/quote]
I got the idea across.

Lucky. PA showed up at our room because we were the loudest, but then they turned out to be kind of dull. One of them actually didn't come, as he was apparetly reading. (They were a one asian team too right? I think I know which was reading...)

[quote="Moldytape"]You know the fortuneteller one where it hands out a small slip with words on it? I forgot what it said though. And not to mention that it was terribly humid playing frisbee outdoors. And remember the time we went to Indiana's room for like a midnight party with like 9 people and Lyndon's (mathcrazed) parents came and we had to hide in the bathroom? Everyone else had to hide under the bed and in the closet. And I think Richard came and pretended to take a shower or something. Then we played ratscrew with like 8 people and watched the end of Kill Bill.[/quote]

dude, the best part of that night was when Fish totally called the fact that there were 20 other people in the room


In the triangle ABC, AB = AC and the bisector of angle B meets AC at the point E. Assume that BC = BE + EA, and find, with proof, what the value of angle A is.

Please help!!! I'm really stuck!!!

what?

now remember to draw a diagram. if im thinking right, we are given ab=ac and bc=ce+ea and drawing a diagram we see that ce+ea=ca
so ab=ac, bc=ca, and we hav an equilateral triangle

so all angles are 60 degrees

The problem states that $ BC\equal{}BE\plus{}EA$, not $ BC\equal{}CE\plus{}EA$...


Provided $n$ positive numbers $x_1, \ldots, x_n$ satisfying $\sum \limits^n_{k=1} x_k = 1.$ For fixed $n$ determine the least constant $t$ satisfying $\sum \limits^n_{k=1} \sqrt{4 \cdot x_k + 1} < t.$

$t = \sqrt(n+4n)$ with closest to t being when there are enormously many x all extremely small.

proof is simple, first show that values are best all equal, then write as $n\sqrt{4/n + 1}$


In a certain kingdom, the king has decided to build 25 new towns on 13 uninhabited islands so that on each island there will be at least one town. Direct ferry connections will be established between any pair of new towns which are on different islands. Determine the least possible number of these connections.

Sorry. Posted before.

http://www.mathlinks.ro/Forum/viewtopic.php?p=190417


Let a,b,c and x,y,z be real positive numbers such that x+y+z=1. Prove that

xa+yb+zc+2 \sqrt (xy+yz+zx)(ab+bc+ca) \leq a+b+c

Source: 2001 UKRAINIAN MATHEMATICAL OLYMPIADS (Grade11)

a + b + c is clearly positive, so divide the equation by k = a + b + c, and let a' = a/k, b' = b/k, c' = c/k.

We then have a' + b' + c' = 1 and we must prove:

xa'+yb'+zc'+2 \sqrt (xy+yz+zx)(a'b'+b'c'+c'a') \leq 1.

We have (\sum xy) = ((\sum x)^2 - (\sum x^2))/2 = (1 - (\sum x^2))/2,
(\sum a'b') = ((\sum a')^2 - (\sum a'^2))/2 = (1 - (\sum a'^2))/2.

By AM-GM and substitution, 2 \sqrt (xy+yz+zx)(a'b'+b'c'+c'a') \leq (xy + yz + zx + a'b' + b'c' + c'a') = 1 - ((\sum a'^2) + (\sum x^2))/2.

By AM-GM, a'x + b'y + c'z \leq (a'^2 + x^2 + b'^2 + y^2 + c'^2 + z^2)/2.

Summing, we obtain LHS \leq 1, which is equal to the RHS.

Equality holds iff x = a', y = b', z = c'.


inequalities
number theory unsolved
number theory

$ S_{1},S_{2}$ denote the sum of the numbers of odd fesp even divisors of all the numbers $ 1,2,...,n$
Prove that      $ | S_{1}-S_{2}-n.\ln2|<\sqrt n+1$
Comment: I think this above problem is well-known(It comes from KVANT magazine).Why number $ 2$ appears in this inequality?

@first: $ |S_{1}\minus{}S_{2}\minus{}n.ln(2)|<\sqrt{n}\plus{}1$
[quote="Nbach"]Why number $ 2$ appears in this inequality?[/quote]
because if $ n \equal{} 2^{a}b$ ($ b\ odd$)$ |$the sum of the numbers of odd divisors of $ n$- the sum of the numbers of even divisors if $ n$$ | \equal{} (a\minus{}1)d(b)$
so $ S_{2}\minus{}S_{1}\equal{}\sum_{b\ odd\in\{1,2,...,n\}}d(b)\frac{a(a\minus{}1)}{2}$ where $ a \equal{} E(ln(\frac{n}{b})\frac{1}{\boxed{ln(2)}})$

[quote]nb: $ d(b) \equal{}$numbre of divisor of $ b$[/quote]


[quote="arqady"]$\sum_{i=1}^n(x_{i}-\frac{n-1}{x_i}+\frac{n^2}{\sqrt{n-1}}(\frac{1}{x_i^2+1}-\frac{1}{n}))\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{i=1}^n\frac{(x_i-\sqrt{n-1})^2(x_i^2-1+\frac{n-2}{\sqrt{n-1}}x_i)}{x_i(x_i^2+1)}\geq0.$[/quote]

Arqady could you explain this step?. How did you go from one to another equivalently?. I make my computations and I didn't find these

[quote="silouan"]Nice solution .My solution is to make the problem as( with some subtitution) as :

Let $k_1,k_2,..k_n$ posiive reals such that $k_1+k_2+...+k_n=1$  and we need to

prove that $\sqrt{\frac{1-k_1}{k_1}}+...+\sqrt{\frac{1-k_n}{k_n}}\geq (n-1)(\sqrt{\frac{k_1}{1-k_1}}+....+\sqrt{\frac{k_n}{1-k_n}})$.

and then it is the problem 52 in ''old and new inequalities'' of Vojtech Jarnic.

But your solution is better than mine.[/quote]
if you have the book of  ''old and new inequalities''  in computer,can you send it to me?

[quote="zhaobin"][quote="e.lopes"][quote="zhaobin"]who can tell me why $x_i \geq \frac{1}{\sqrt{n-1}}$
thanks[/quote]

because $x_i \geq 1$  ;)[/quote]
yes,you are righyt.
but  can the inequality be ture without the condition $x_i \geq 1$?[/quote]

i think that the silouan's solution do not depend of this.  :)

[quote="silouan"]Yes the ineq holds without  this. My solution proved this. See above.[/quote]
where?i didn't have the book.

[quote="silouan"]I have not the book in the computer so I will post the solution. This

$\sqrt{\frac{1-k_1}{k_1}}+...+\sqrt{\frac{1-k_n}{k_n}}\geq (n-1)(\sqrt{\frac{k_1}{1-k_1}}+....+\sqrt{\frac{k_n}{1-k_n}})$ is equivalent to show that

$n\sum_{i=1}^n {\sqrt{\frac{k_i}{1-k_i}}\leq (\sum_{i=1}^n k_i)(\sum_{i=1}^n {\frac{1}{\sqrt{a_i(1-a_i)}}})}$.

But the last comes from Chebychev's inequality[/quote]
oh,i know,thanks.
but i think $a_i=k_i$


prove that

1/sin1sin2 + 1/sin2sin3 + 1/sin3sin4 + ... +1/sin89sin90 = cos1/sin^2(1)

everything is in degrees.

hey buddy, its me again

here is your solution 1/sinxsin(x+1)= sin(x+1-x)/sinxsin(x+1)sin1

={sin(x+1)cosx-cos(x+1)sinx}/sinxsin(x+1)sin1
=1/sin1{cotx-cot(x+1)}

now sum up x=1 to 89

1/sin1{cot1-cot90}=1/sin1{cos1/sin1}

hence the result follows


This can get locked after it's served it's purpose, but is it possible for me to get a hotel near the one MathCounts has booked and go inside and meet people?

-Jorian

Jorian, you can stay in the same hotel and eat all the meals with the kids that made it.  You can usually attend some of the special events too.

You can even register online right here.

[url]https://www.ptfassociates.com/secure/mathcounts/Registration.htm[/url]

Frost

Ok, thank you very much!!!

I couldn't find any rooms when I looked at the Hilton website a month ago or so.

so this will help

Thanks,
Jorian

edit: (so don't lock this yet!)

wait, one problem, i don't see the price for the hotel anywhere, just the meals and the competition thingys

jorian

I just realized that the hotel listed on that page is WRONG.  That is the hotel in Arlington, VA from last year.  The correct hotel is :

Fort Worth Convention Center and Renaissance Worthington Hotel
Fort Worth, Texas

You have to get your hotel rooms on your own, but when you call tell them you are a guest with the mathcounts competition.  You might even get a special rate!

shoot, i e-mailed mathcounts with that link!

there's no rooms left though for that hotel!

jorian

Really? Then how are our family members going to...? Are they staying in the same hotel as we are?

I e-mailed mathcounts with the following message:

[quote=Jorian e-mailed MathCounts...]
The last e-mail I gave had a wrong link. Please still read it as it includes other questions of mine.

Could you provide me with the information to possibly going to nationals this year?

I looked at the Hilton website some time ago and it appeared as the rooms were booked, but possibly you have some at special rates. I would need to pay for the transportation to and from the airport as I’m only 13 years old. I would REALLY REALLY REALLY like to experience nationals and be with the people there as I know quite a few that are definitely going (they are really good) and ones that just made it today.

Thank you soooooooo much!

Jorian Hoover
[/quote]

I think they booked the whole hotel, and they have extras rooms. I think my message is good.

-jorian

MC already reserved rooms for MC parents.

How about rooms for people like me?

-jorian

No, parents of competitors have first priority, and there are never any rooms leftover.

Maybe you can get a hotel across the street or something...  And then just show up and say you're someone's sibling or something... :D

On a side note, I cant wait to meet a lot of the people on this forum...

[quote="jhredsox"]I hope you're kidding.

Didn't fanatic go last year though?

-jorian[/quote]

No, my parents almost didn't get a room in March last year from the reserved rooms.

Fanatic did go! he told people he was going and i saw him right after he came back at like a ball state award ceremony at Indiana Academy.

You can go without being in the same hotel!!!

[quote="frost13"][quote="jhredsox"]Ouch, that's expensive. So the hotel's like 300 dollars a night at that rate?

-jorian[/quote]

No.  Did you actually look at the registration form.  Food is like $\$$250, all tolled on that registration form is close to $\$$350.  Flights were like $\$$500.  That's $\$$850 before the hotel.  The hotel, if you get in as a guest, will be at a discounted Mathcounts rate.

Can we end this talk and just call tomorrow and find out.  Then you can tell us all the real info![/quote]
Eh, it's much much cheaper not to purchase meals with them, 16 meals can be reduced to just 80 bucks for 5 bucks a meal.

[quote="Fanatic"]If you're in 7th grade:

You know what's coming up. It's decently fun [WHY DOES IT HAVE TO BE IN [b] TEXAS??? [/b]... i mean nothing against texas... at least be in like dalls, san antonio, or houston... WHY DOES IT HAVE TO BE IN [b] Ft. WORTH???[/b]...]. You will know what to expect for 8th. It makes things so much sweeter once you actually make it to nats.[/quote]

Fort Worth is close to Dallas...  I would like it if it was in Houston.  Then it would really be full circle since I was born around Houston...

It would be awesome if Nationals was in Houston, which is lk an hour away from here. Of course I'm also kind of disappointed that nationals is in Texas. I really wanted to go visit some cool city far away.

Of course this is assuming I make nationals. :|

Ok, today I'm home sick :( . Apparently they do have some rooms on the website, but the cheapest one is $\$$350 a night (they booked the two types in the $\$$200's). I would call, but my mom has put the calling card for long distances somewhere and I can't find it. I was also wondering if anyone would like to stay in the same room as me (possibly a student who also wanted to be at natns).

And I have to go. When else are most of you going to be under the same roof?

But seriously, does someone else want to go?

Thanks,
Jorian

spam, i need to meet people from aops.

plus, i thought you wanted to meet me...  :huh:

-jorian

that weren't spam, I just had some LaTeX booboos

[quote="kstan013"]A. Your school won't let you skip
B. $\$$
C. Your parents won't let you go (or skip school)
D. $\$$
E. It's more meaningful when you actually EARN your way there
F. Did I mention the money it would cost?[/quote]

let me respond in that case

A. yes they will
B. got it covered
C. yes they will
D. got it covered
E. i know, i just have to meet aopsers
F. no you didn't, your latex errors were hilarious though

ok, but seriously, i need someone else to come with me

jorian

[quote="Fanatic"]If you're in 7th grade:

You know what's coming up. It's decently fun [WHY DOES IT HAVE TO BE IN [b] TEXAS??? [/b]... i mean nothing against texas... at least be in like dalls, san antonio, or houston... WHY DOES IT HAVE TO BE IN [b] Ft. WORTH???[/b]...]. You will know what to expect for 8th. It makes things so much sweeter once you actually make it to nats.[/quote]

or you could watch it on ESPN2 :D

are they gonna have it this year???

jorian

They didn't last year, so I don't really think so.

you're pretty close with Tobias's parents right? you could just ask them to record it and make a copy for you.

hell no, i'm meeting aopsers therer. video doesn't cover it!!!

please read my topic that i made not too long ago as it is nicely written. don't spam on there please as it should be used for info purposes.

jorian


Let $a_1$, $a_2$, ..., $a_{10}$ be positive integers such that $a_1<a_2<...<a_{10}$. For every $k$, denote by $b_k$ the greatest divisor of $a_k$ such that $b_k<a_k$. Assume that $b_1>b_2>...>b_{10}$. Show that $a_{10}>500$.

Let $a_k=b_kc_k$. Then we have $a_{10}>\cdots>a_1$ and $c_{10}>\cdots>c_1$, and $c_k$ is the least prime divisor of $a_k$. $a_{10}$ must be composite, so $a_{10}=p_{10}q_{10}$ and let $p_{10}\geq q_{10}$. If $a_{10}\leq500$, then $q_{10}\leq\left[\sqrt{500}\right]=22$ thus the least prime divisor of $a_{10}$ (that's $c_{10}$) is $\leq 22$. But then $c_{10}>\cdots>c_1$ isn't possible, because $c_k$ are all primes. ($c_1$ may be $1$)

Let $p_i$ be the least prime divisor of $a_i.$ Since, $a_i < a_{i+1},$ we must have that $p_{i} > p_{i+1},$ as note that $b_i = a_i / p_i$ and $b_i > b_{i+1}.$ So no two of $a_1,\dots,a_{10}$ have the same least prime divisor. So, $p_1 < p_2 < \dots < p_{10}.$ Obviously, none of $a_k, k \in \{1,\dots,9\}$ can be primes, as then $b_k = 1,$ and $b_{k+1} < b_{k} = 1.$ Thus, in the minimum case, we assign the 9 least primes values to $p_1,\dots,p_9.$ So, $p_9 = 23.$ Since, $a_9$ cannot be prime, $a_9 \ge 23^2 = 529 > 500.$ Hence, $500 < a_9 < a_{10}.$

Hmm, so would a stronger version of this problem be: Prove that $a_9 > 500.$ ?

Unless I'm making some mistake of course...  :maybe:

To previous post, why must $a_{10}$ be composite?

we call the smallest prime divisor of a(i) p(i) then b(i)=a(i)/p(i) so p(1)<p(2)<p(3)<p(4)<...<p(10)
now we have two situations
1.a(10) is prime
2.a(10) is not prime
it's easy to proof that a(10) is more than 500 in both situations!!!


calculus
integration
calculus computations

Can someone compute \[ \int_0^\infty dx x^{\minus{}3/2} e^{\minus{}\frac{xa^2\plus{}b^2/x}{2}}\]

Use the formula $ \int_0^\infty \exp\left(\minus{}\alpha^2 t^2\minus{}\frac{\beta^2}{t^2}\right) \, dt\equal{}\frac{\sqrt{\pi}}{2\alpha}\, e^{\minus{}2\alpha\beta}$

and differentiate with respect to $ \beta$ :

$ \int_0^\infty \exp\left(\minus{}\alpha^2 t^2\minus{}\frac{\beta^2}{t^2}\right) \,\frac{\minus{}2\beta}{t^2} dt\equal{}\minus{}\sqrt{\pi}\, e^{\minus{}2\alpha\beta}$

Divide both sides by $ \minus{}\beta$ and substitute $ t\equal{}\sqrt{x}$ .

$ \int_0^\infty \exp\left(\minus{}\alpha^2 x\minus{}\frac{\beta^2}{x}\right) \, x^{\minus{}3/2}\, dx\equal{}\frac{\sqrt{\pi}}{\beta}\,e^{\minus{}2\alpha\beta}$

Finally substitute $ \alpha\equal{}\frac{a}{\sqrt{2}}$ and $ \beta\equal{}\frac{b}{\sqrt{2}}$ to obtain

$ \int_0^\infty \exp\left(\minus{}\frac{a^2 x\plus{}b^2/x}{2}\right)\, x^{\minus{}3/2}\, dx\equal{}\frac{\sqrt{2\pi}}{b}\, e^{\minus{}ab}$ .

[quote="gauss202"][quote="Doe John"]Use the formula $ \int_0^\infty \exp\left( \minus{} \alpha^2 t^2 \minus{} \frac {\beta^2}{t^2}\right) \, dt \equal{} \frac {\sqrt {\pi}}{2\alpha}\, e^{ \minus{} 2\alpha\beta}$
[/quote]

Is this a well known formula for you, or did you derive it specifically for this problem?[/quote]

that depends.

i asked this question to a couple of 7th and 8th graders (11-13 yr olds), they failed to show it, but most of 9th and 10 graders did provide the answer. here is the answer that i eventually posted.

let
$ \Omega \equal{} \int_0^\infty\exp\left( \minus{} \alpha^2 t^2 \minus{} \frac {\beta^2}{t^2}\right) \, dt$

$ \Rightarrow \dfrac{d\Omega}{d\beta} \equal{} \int_0^\infty\exp\left( \minus{} \alpha^2 t^2 \minus{} \frac {\beta^2}{t^2}\right)\left( \minus{} \dfrac{2\beta}{t^2} \right) \, dt$

let $ u \equal{} \dfrac{\beta}{\alpha\,t}$ so $ du \equal{} \minus{} \dfrac{\beta}{\alpha\,t^2} \, dt$

$ \therefore \dfrac{d\Omega}{d\beta} \equal{} \minus{} 2\,\alpha \int_0^\infty\exp\left( \minus{} \alpha^2\,u^2 \minus{} \frac {\beta^2}{u^2}\right) \, du$

i.e. $ \dfrac{d\Omega}{d\beta} \equal{} \minus{} 2\,\alpha\, \Omega$

which when solved gives $ \Omega \equal{} C\;\exp\left( \minus{} 2\alpha \beta\right)$

and when $ \beta \equal{} 0$, $ \Omega \equal{} C \equal{}\int_0^\infty \exp\left( \minus{} \alpha^2 t^2  \right) \, dt \equal{} \dfrac{\sqrt\pi}{2\alpha}\quad\Rightarrow \quad \Omega \equal{} \dfrac{\sqrt\pi}{2\alpha}\;\exp\left( \minus{} 2\alpha \beta\right)$

[quote="gauss202"][quote="Doe John"]Use the formula $ \int_0^\infty \exp\left( \minus{} \alpha^2 t^2 \minus{} \frac {\beta^2}{t^2}\right) \, dt \equal{} \frac {\sqrt {\pi}}{2\alpha}\, e^{ \minus{} 2\alpha\beta}$
[/quote]

Is this a well known formula for you, or did you derive it specifically for this problem?[/quote]

This formula is Laplace Transform. $ \left(\text{Case by} \int_0^\infty e^{\minus{}ax^2}F(x)dx \equal{} \int_0^\infty e^{\minus{}at}\dfrac{F(\sqrt{t})}{\sqrt{t}}dt \right)$
$ \int_0^\infty \exp \left[ \minus{}c\left( \dfrac{x^2}{a^2} \plus{} \dfrac{b^2}{x^2} \right) \right] dx \equal{} \dfrac{a}{2}\sqrt{\dfrac{\pi}{c}}e^{\minus{}\dfrac{2bc}{a}} \ \ (a, b, c>0)$

Another Laplace Transform. $ \left(\text{Case by} \int_0^\infty e^{\minus{}ax}F(x)dx\right)$
$ \int_0^\infty e^{\minus{}ax} \cdot \dfrac{e^{\minus{}\dfrac{b^2}{x}}}{\sqrt{x}}dx \equal{} \sqrt{\dfrac{\pi}{a}}e^{\minus{}2b\sqrt{a}} \ \ (a, b>0)$

I convinced myself with a primitive in terms of the erf(x) function. Try it with the integrator

@misan and others wanting to comment on misan's claim:
[url=http://www.mathlinks.ro/viewtopic.php?t=194271]here[/url]we are having a discussion about this topic in the more suitable indian forum..please do join us :) ...

Moderator edit: No more discussion of misan's claim here, please. All previous posts concerning it are now moved to the India forum. You can find them [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194512]there[/url]. The thread is unlocked but only for the discussion of the problem itself!


Let be given $n$ real numbers: $x_1\ge x_2\ge x_3\ge ....\gex_n$.Let $S=\sqrt{\frac{\sum_{1\le i<j\le n}(x_i-x_j)^2}{n-1}}$.
Prove that:
$na_n\le \sum_{i=1}^n x_i-S$
$na_1\ge \sum_{i=1}^n x_i+S$


Hi,

Your inequality $ \frac {|x \plus{} 4|}{|x|} \leq 3$ is equivalent with $ |x \plus{} 4| \minus{} 3|x|\leq 0$.

Now just check the intervals $ x\in( \minus{} \infty , \minus{} 4] ,[ \minus{} 4,0)$ and $ (0,\infty)$

I believe you mean the intervals $ ( \minus{} \infty , \minus{} 4] ,[ \minus{} 4,0)$and$ (0,\infty)$.

I think it should be like this:

(I).$ \frac {x \plus{} 4}{x}\le3$ we get $ x\ge 2$

(II).$ \minus{} \frac {x \plus{} 4}{x}\ge3$ we get $ x\le \minus{} 1$

So the solution is  $ x\in (\minus{}\infty,\minus{}1]$ and $ x \in [2,\infty)$

but the condition is x belonging to [-4,0]

so my soln..........[-4,-1] and [2,infintiy]

pl tell rigth or wrong

I also got $ x\in (\minus{}\infty,\minus{}1]$ and $ x\in [2,\infty)$

@[b]Arnav...!!![/b], the condition isn't implied in the equation. Also $ x \ne 0$. Otherwise the equation would be undefined.

[quote="Elio (n)"]Solve the inequation : $ \frac {|x \plus{} 4|}{|x|}\le3$[/quote]

Simply note that it's equivalent to $ \left|\frac{x\plus{}4}{x}\right|\le 3$, and you get $ \frac{x\plus{}4}{x}\le \pm 3$, from which the answer follows.

$ \boxed{x\le \minus{}1\text{ or }x\ge 2}$


Ah sorry, this is my proposed problem for challenged contest in mathnfriend, so please don't discuss it now my friends  :) .

[quote="Anh Cuong"]Since the above problem we shouldn't discuss it now, maybe everybody want to try this:
Give $a,b,c \geq 0$. Prove that:
$2(a^{5}+b^{5}+c^{5})+(a^{4}b+b^{4}c+c^{4}a) \geq 3(ab^{4}+bc^{4}+ca^{4})$
Have fun  :) .[/quote]

Hey, for the general case of your inequality, I proved that
$k(a^{5}+b^{5}+c^{5})+l(a^{4}b+b^{4}c+c^{4}a) \geq (k+l) (ab^{4}+bc^{4}+ca^{4})$
when $a,b,c \geq 0$, and $k>0$
  Is it also your answer? :)

You mean this problem:
$(a^{5}+b^{5}+c^{5})+k(a^{4}b+b^{4}c+c^{4}a) \geq (k+1)(ab^{4}+bc^{4}+ca^{4})$
This one is not valid for all $k$ then  :( .

[quote="Gibbenergy"]
Hey, for the general case of your inequality, I proved that
$k(a^{5}+b^{5}+c^{5})+l(a^{4}b+b^{4}c+c^{4}a) \geq (k+l) (ab^{4}+bc^{4}+ca^{4})$
when $a,b,c \geq 0$, and $k>0$
  Is it also your answer? :)[/quote]

I check my proof again and I think that it is ok. The only assumption that I doubted is assume $a \geq b \geq c \geq0$
Here is my solution:
Let $f(a) =k(a^{5}+b^{5}+c^{5})+l(a^{4}b+b^{4}c+c^{4}a)-(k+l) (ab^{4}+bc^{4}+ca^{4})$
$f'(a)=5ka^{4}+l(4a^{3}b+c^{4})-(k+l)(b^{4}+4a^{3}c)$
$f''(a)=20ka^{3}+12la^{2}b-12(k+l)a^{2}c =4a^{2}(3k(a-c)+3l(b-c)+2ka)$
if $k,l \geq 0$ and $a\geq b \geq c$ then $f''(a) \geq 0$
then $f'(a) \geq f'(b)=(b-c)(l(b^{3}-c^{3})+b^{3}l-b^{2}cl+b^{3}l-bc^{2}l+4b^{3}k))  \geq 0$
so $f(a) \geq f(b)=k(b+c)(b^{2}+c^{2})(b-c)^{2}\geq 0$

In the cyclo if we assume $c=min{a,b,c}$ then this aproach works well, right? So I think that if $k,l >0$ then the general inequality will be corrected.
 :lol:

[quote="Gibbenergy"] The only assumption that I doubted is assume $a \geq b \geq c \geq0$
[/quote]
You correctly doubt! What if $b\geq a\geq c$ $?$

Indeed, for $a \geq b \geq c$ we have: $a^{4}b+b^{4}c+c^{4}a \geq ab^{4}+bc^{4}+ca^{4}$ but when $b \geq a \geq c$ the sign is reversed.


induction
trigonometry
geometry
geometric transformation
homothety
number theory
greatest common divisor

Tomorrow night, I'll post a PDF of the second level Olympiad.  Please wait to discuss it until then.  I'll post the solutions a week from today.  (If you post a solution that's very different from my original solution, I'll add it to the official solutions and give you credit.)

Thanks.

Rick Yanco
Worcester Academy

Yesterday's contest is now online at http://www.matharchive.org/olympiad/index.html.  (Click "Two" in the first line)

My thoughts:

[hide="1"] (a) and (b) were just computation. (c) was a little tricky because (If I calculated right) it appears that the hands will be perpendicular twice per revolution of the minute hand, but actually on two revs it's only once. (d) wasn't bad as it looked, the hour hands are always parallel so it's just finding when the minute hands match up, which is straightforward after you compute their angular velocities.
[/hide]

[hide="2"] Again, (a) and (b) were pretty straightforward. The recursion was just $ f(n) \equal{} f(n\minus{}2) \plus{} f(n\minus{}3)$, and you prove this using the standard method: either the first seat is occupied or the second one is.

[/hide]

[hide="3"] This one was very straightforward, just algebra. [/hide]

[hide="4"] For (a) just combine the sums to a product. (b) is a little tricky; you need to do some algebra manipulation, and you realize that $ x\equal{}(2n\plus{}1)^{n\plus{}1}, y\equal{} (2n\plus{}1)^n$ gives an infinite family.
[/hide]

[hide="5"] Induction and sum-to-product formula for $ \sin$ kills it. [/hide]

[hide="6"] Don't know if there's a better way to do this one, but if you set $ OA\equal{}R$, $ PJ \equal{} r$, $ OC\equal{}x$, the answer follows from Pythagorean and some algebraic manipulation. [/hide]

[hide="7"] (a) was standard, (b) was Derangements. I ran out of time bruteforcing PIE on (c), which looked pretty awful. [/hide]

[quote="Phelpedo"]Don't know if there's a better way to do this one, but if you set $ OA \equal{} R$, $ PJ \equal{} r$, $ OC \equal{} x$, the answer follows from Pythagorean and some algebraic manipulation.[/quote]

That's what I did, but you can note that there is a homothety centered at the point of tangency, an easier solution follows.

How did you approach proving that you get a primitive triple in 3d? Maybe I'm missing something obvious.

I managed to make stupid mistakes on 1a, 1c, 3c (wrote down the wrong recursion, wtf), 3d (because of 3c), 3e (managed to prove the wrong recursion by messing up notation), 7c.

You had to show that GCD[a+2b+2c, 2a+b+2c, 2a+2b+3c] = 1. Letting a+b+c=P (it is, after all, a triangle), we can write it cleanly as:

GCD[2P-a, 2P-b, 2P+c]
= GCD[(2P-a) - (2P+c), (2P-b) - (2P+c), 2P+c]
= GCD[-a-c, -b-c, 2a+2b+4c-c]
= GCD[a+c, b+c, 2(a+c)+2(b+c)-c]
= GCD[a+c, b+c, -c]
= GCD[a+c, b+c, c]
= GCD[a, b, c]
= 1

Sweet. I guess that would have been better than doing 2 at a time.

[hide="1."]
(a) I messed up on due to thinking it was 17 minutes AFTER 17:17

(c) I solved this completely algebraically - used mod 720

(d) similar as (c)
[/hide]

[hide="3."]
This was mostly simple algebra.

For the primitive part, I just used Euclidean algorithm on two numbers as follows:

GCD[a+2b+2c,2a+2b+3c]=...=GCD[2b-a,a+c]

Assume GCD[2b-a,a+c]>1. Then there exists a prime p such that p|2b-a,a+c. Thus $ p|(a \plus{} c)(c \minus{} a) \equal{} c^2 \minus{} a^2 \equal{} b^2$ so $ p|b$. But then since $ p|2b \minus{} a$, we have $ p|a,c$, which is a contradiction, so the gcd is $ 1$.

I also proved (d) first. Then I noted that [a b c]B = [a -b c]A and since a,-b,c is still a Pythagorean triple, I cited (d) to prove (c) ii. Then I used a similar idea for the other parts.
[/hide]

[hide="5."]

This was really simple using induction. Just prove part (e) first.
[/hide]

[hide="6."]

I did the same as Phelpedo.

[/hide]

A note on number 3:

[url=http://www.cut-the-knot.org/pythagoras/PT_matrix.shtml]Here is one website[/url] discussing the theory behind the matrices.

For general comments:

[hide="1"]Some computation. I might've made a few careless errors here.[/hide]
[hide="2"]A standard problem goes: you walk up a staircase, and at each move, you can either go to the next step or skip a step. How many ways to get to the top? The answer is the Fibonacci series. This problem had a very similar underlying theory. For the purposes of the recursion, one can show that it is the same if you assume the first and last seats are filled, which makes analogous reasoning much easier.[/hide]
[hide="3"]This problem has deep roots, but it seems that in an effort to water it down to something doable within the timeframe, much of this depth is lost, although the problem setup is still highly suggestive of the general result (see the link above). Part d was the interesting part because showing that the resulting triple is primitive involved some manipulation.[/hide]
[hide="4"]IMHO this problem was overvalued pointwise. Part a was an application of the log chain rule and the log product rule, and was straightforward (for anyone who noticed the factorial sign -- many people did not).

Part b looked interesting, but performing $ x^{x\minus{}y} \equal{} y^{x\plus{}y} \iff \left( \frac{x}{y} \right) ^x \equal{} (xy)^y$ made the substitution $ x\equal{}ay$ pretty clear, and since we only had to show existence, one could just solve for $ y$ in terms of $ a$, and then $ x$ would be $ ay$, giving an infinite family.[/hide]
[hide="5"]This was easy to put away using induction with sum/product formulas or by re-writing it in exponential form and cranking out the geometric series.[/hide]
[hide="6"]I didn't have time to do this one, and I regret that. If I had time, I would have solved it along the lines of what Phelpedo suggested[/hide]
[hide="7"]a and b were standard, though I randomly forgot the derangements formula and tried to crank out a recursion, and I think I messed it up.

I'm curious to see a nice way of doing c.  :D [/hide]

Someone at my school came up with this for (c)

Consider the colors 1,2,...,n, where each color only matches with color+1 or color-1 if it exists. Let $ P_{n}$ denote the probability he gets through $ n$ days.

Consider what one of the 1's is paired up with. It is paired up with the other 1 with probability $ \dfrac{1}{2n \minus{} 1}$, then the probability that the rest is OK is $ P_{n \minus{} 1}$. It is paired up with one of the 2's with probability $ \dfrac{2}{2n \minus{} 1}$, but then the other 1 has to be paired up with the other 2, with probability $ \dfrac{1}{2n \minus{} 3}$. Then, there's a $ P_{n \minus{} 2}$ that the rest works.

So we have the recursion $ P_{n} \equal{} \dfrac{P_{n \minus{} 1}}{2n \minus{} 1} \plus{} \dfrac{2P_{n \minus{} 2}}{(2n \minus{} 1)(2n \minus{} 3)}$.

I'm a bit confused -- doesn't each color also match with color+0?  :maybe: Otherwise, interesting solution

[quote="The Zuton Force"]I'm a bit confused -- doesn't each color also match with color+0?[/quote]

if i understand what you mean, isn't that covered in the first case? when 1 is matched with 1, and you do whatever with 2, 3, ..., n?

I found 7 part c much easier than part b, did I miss a really simple solution to it?

Also, do we have answers?

My (fairly official) solutions are online: http://www.matharchive.org/olympiad/index.html.  (Click "Solutions" under "Two" in the first line.)

I added a short note about the study hall problem, and a lengthier note about the primitive triples problem.

Enjoy.

Ahh.. made arithmetic error on 7b. :huh:
My solution to 7c using a tree for smaller numbers:

call colors A, B, C, D, E.
Arrange all pairings in alphabetical order.

The first element must be A.
The 2nd element has a 1/9 chance of being an A.
-The 3rd element must then be a B.
-The 4th element has a 1/7 chance of being a B.
--The 5th element must then be C.
--The 6th element has a 1/5 chance of being C.
---Remaining elements do not matter
--The 6th element has a 2/5 chance of being D.
---7th is C.
---8th has 1/3 chance of being D
---Remaining do not matter.
--1/7 * (1/5 + (2/5 * 1/3)) = 1/21
-The 4th element has a 2/7 chance of being C.
--The 5th element is B.
--The 6th element has a 1/5 chance of being C.
---Remaining do not matter.
--2/7 * 1/5 = 2/35
-1/9 * (2/35 + 1/21) = 11/945
The 2nd element has a 2/9 chance of being a B.
-3rd is A.
-4th has 1/7 chance of being B.
--same as before.
-1/21 * 2/9 = 2/189
11/945 + 2/189 = 21/945 = 1/45.


AMC
USA(J)MO
USAMO
Putnam
inequalities
AIME
analytic geometry

The people who compete are extremely smart.  It doesn't matter that they prepare, the fact that they can think that far out of the box is not something that can be taught.

The IMO is nothing special. From our oh-will-I-make-USAMO viewpoint, it is, but their are harder national competitions, especially in eastern Asia. And it's completely preparation. It doesn't matter how utterly brilliant person N is, if person N doesn't know theorems X, Y, and Z, then he gets a 0 on the IMO like many people there. There's no such thing as "natural" intelligence or creativity. Everything is preparation; if they were taught how to problem solve, then good for them.

If you can come across some book that has every theorem necessary and how to apply them, please share with the world.  The AoPS books don't guarantee you'll make MOP.  The people who do well on Putnam are very smart, it has little to do with training.

[quote="Temperal"]It doesn't matter how utterly brilliant person N is, if person N doesn't know theorems X, Y, and Z, then he gets a 0 on the IMO like many people there.[/quote]

Well, if you had NO knowledge whatsoever, then I'd understand, but good IMO problems usually just use clever applications of very basic ideas.  (for example, I would not consider #6 this year to be a good IMO problem, because from what I could see, there was no really elementary way to solve it.  #3, on the other hand, was really really nice)
So then I suppose it depends on how good the problems are.  I know the USAMO generally has nice problems.

Some problems, like number 3 this year, only require intelligence. They are extremely beautiful - and just as rare.
The vaste majority of the problems require more training than intelligence. Which does not mean anyone can solve them after a sufficient amount of training, of course.

JRav: ...your post an a cover-all book with all theorems has little/nothing to do with the topic. I never said such a book existed. What does it have to do with anything?

K81o7: Now, here's an example of an actual intelligent objection. You seem to be assuming creativity and problem-solving skills are "natural". Why can't they be taught; and thus be the source of this apparently useful resource on the IMO?

Of course, to have creativity is certainly helpful. However, I believe that training is just as, probably more, important. Learning the 'tricks' of the trade such as the numerous theorems on inequality posted on this forum is very helpful. Also the use of 'Vieta Jumping' on problem 5 this year is another one of these 'tricks'. Being able to think critically about problems (such as no. 3 IMO 2007) will help in many cases.

If you don't know certain theorems then you may have little chance of solving some problems (no.6 IMO 2007). Also, recall how the converse of the problem no.2 this year was seen before so this shows that trying a variety of problems is essential.

[quote="FOURRIER"][quote="moogra"]I've always thought problems on the AIME and stuff have a trick to help. Of course brute force methods always help as well.[/quote]

What is [b][u]brute force[/u][/b] please?

I've heard it thousands of times  :P[/quote]

Using coordinates to solve a geometry problem

Given an equation of some sort, checking all solutions by hand

Given a counting problem, counting all cases by hand

etc etc

I'd say you would probably have to be both;If you're extremely smart, but don't prepare at all, you will definitely fall behind. On the other hand, if you're willing to work hard, but have no sense of creativity, you won't do that great. It's best to be naturally smart, but to also work hard.

[quote="Peter"]
The vaste majority of the problems require more training than intelligence. [/quote]

Yes, Unfortuntly   :(

[quote="K81o7"]
Using coordinates to solve a geometry problem

Given an equation of some sort, checking all solutions by hand

Given a counting problem, counting all cases by hand

etc etc[/quote]

Now I see :D , Thanks  :wink:

Is there any brute force for inequalitites  :P ?

Or what we did at MOP...Dumbass that inequality/expand that polynomial.  In other words, the most elegant form of inequailty bash is expanding out and using Schur (in the cases where you can).

But yeah...calculus is the real brute force way.  (Then again, I wrote a very elegant solution to an inequality at MOP taking a derivative once...)


Oh, wow... the proof in the book is beastly.  I hope someone can come up with a more intuitive proof than this one :|

[hide="Engels one-liner"]
$\left( \sum_{i=0}^{n} a_i \right)^2 - \sum_{i=0}^{n} a_i^3 = 2 \sum_{i=0}^{n} \sum_{j=0}^{i} a_i \frac{a_j + a_{j-1} }{2} [ 1 - (a_j - a_{j-1} ) ] \ge 0$
[/hide]

[quote="probability1.01"]Induction will probably work.[/quote]

[hide]
You are right, induction works. Actually, we can prove the following first (by induction or whatever method):

$\sum_{i=0}^na_i\geq \frac{1}{2}a_na_{n+1}$
[/hide]