number theory
least common multiple
greatest common divisor
MATHCOUNTS
relatively prime

what is the greatest integer than cannot be obtained by addining non negative multiples of 6, 10 , and 15?

is there a quick formula for questions like these?

I think a fast way would be two recognize 6=2*3,10=5*2,15=3*5. But from here i am stuck, alternative-brute force it.

considering units digits can for some reason make it easier to find out

6, 10, and 15.
Well, using 6 and 10, the largest we can form is infinitely large.
However, we can form all even numbers larger than 14.
Now consider if we have 15. Then, for all odd numbers we need one 15 and then some number of 6 and 10. However, 14+15 would be impossible and the largest number we can't form.
14+15 = 30, just kidding, 29.

15+ (any even number>15)=any odd number>30, and we already got all even numbers, so 29.

Yes, it is a coincidence.

I think there was a formula posted here sometime for 3 values, but I forgot it and I wasn't sure if it was correct in the first place.

[quote="uldivad9"]Well, their LCM has to be one, otherwise the problem is unsolvable... if they had a LCM of n, you would not be able to make any number congruent to mn-(n-q) for positive integers n, m, and q, q < n.[/quote]

I think you mean GCF :wink:

[quote="unimpossible"]any coincidence that its their LCM-1?[/quote]

Hey, that's what he said, and I forgot to correct it. You know what I mean. :P

[quote="unimpossible"]any coincidence that its their LCM-1?[/quote]
Yes, it's a coincidence.  A quick counterexample... Consider the three summands, $ 15,$ $ 21,$ and $ 35,$ which are pairwise products of three distinct primes, as in the original problem.  I assert the largest number that cannot be achieved is $ 139.$

First, let's assume you know the Chicken McNugget Theorem, which says if you have exactly [b]two[/b] relatively prime positive integers, $ s$ and $ t$, the largest number that cannot be achieved through summation is their product minus their sum, $ (s\times t) \minus{} (s \plus{} t).$  So, if you can buy McNuggets only in cartons of 5 or 7, the largest number you cannot purchase is $ 35 \minus{} 12 \equal{} 23.$

Let's consider how we might achieve the number $ 139,$ above.  Let's repeatedly take away $ 35$, and consider what remains.  $ 139,$ $ 104,$ $ 69,$ and $ 34.$  We must use 15's and 21's to sum to one of these numbers.  But 15 and 21 are multiples of 3.  The only multiple of 3 in our list is 69.  But 69 is the largest number you cannot achieve with 15 and 21.  (Divide everything by three.  23 is the largest number you cannot achieve with 5 and 7.)

Let's illustrate a different way...  We'll take away $ 21,$ and consider what remains.  $ 139,$ $ 118,$ $ 97,$ $ 76,$ $ 55,$ $ 34,$ and $ 13.$  We must use 15's and 35's to sum to one of these numbers.  But 15 and 35 are multiples of 5.  The only multiple of 5 in our list is 55.  But 55 is the largest number you cannot achieve with 15 and 35.  (Divide everything by five.  11 is the largest number you cannot achieve with 3 and 7.  $ 21 \minus{} 10 \equal{} 11.$

Let's try a third way...  We'll take away $ 15,$ and consider what remains.  $ 139,$ $ 124,$ $ 109,$ $ 94,$ $ 79,$ $ 64,$ $ 49,$ $ 34,$ $ 19,$ and $ 4.$  We must use 21's and 35's to sum to one of these numbers.  But 21 and 35 are multiples of 7.  The only multiple of 7 in our list is 49.  But 49 is the largest number you cannot achieve with 21 and 35.  (Divide everything by seven.  7 is the largest number you cannot achieve with 3 and 5.  $ 15 \minus{} 8 \equal{} 7.$

All three of these give us good patterns to work with.  (Always look for patterns with MATHCOUNTS.)  I will assert (but not prove) the formula for this special case:

[hide]Given three distinct primes, $ p,$ $ q,$ and $ r,$ taken pairwise to make three summands, $ pq,$ $ qr,$ and $ rp\ldots$

This formula follows directly from any of the above three examples:
$ ((p\times q) \minus{} (p \plus{} q))\times r \plus{} (p\times q)\times(r \minus{} 1)$

Simplify...
$ pqr \minus{} pr \minus{} qr \plus{} pqr \minus{} pq$

$ \boxed{2pqr \minus{} (pq \plus{} qr \plus{} rp)}$

For this problem, $ 2\times 3\cdot 5\cdot 7 \minus{} (3\cdot 5 \plus{} 5\cdot 7 \plus{} 7\cdot 3) \equal{} 2\times 105 \minus{} (15 \plus{} 35 \plus{} 21) \equal{} 210 \minus{} 71 \equal{} 139$

For the original problem, $ 2\times 2\cdot 3\cdot 5 \minus{} (2\cdot 3 \plus{} 3\cdot 5 \plus{} 5\cdot 2) \equal{} 2\times 30 \minus{} (6 \plus{} 15 \plus{} 10) \equal{} 60 \minus{} 31 \equal{} 29$[/hide]

Problems involving two numbers are common in MATHCOUNTS, and high-level competitors should know the Chicken McNugget Theorem.  Problems involving more than two numbers are uncommon in MATHCOUNTS.  Such problems will almost surely be small enough to use brute force.

A perfectly fair, State- or National-level problem:  What is the largest sum that cannot be attained by adding non-negative multiples of 6, 14, and 21.

what if you fitted the numbers 8,14,19 would it be 39 then?


integration
real analysis
real analysis unsolved

Let $ a_1,a_2,..,a_n$ are $ n$ real number .
Calculate
 $ \int_{0}^{1}\int_{0}^{1}{(x\plus{}a_1y)(x\plus{}a_2y)...(x\plus{}a_ny)dxdy}$

Well, $ \prod_i (x\plus{}\alpha_i y) \equal{} x^n \plus{} x^{n\minus{}1}y\sum_i \alpha_i \plus{} x^{n\minus{}2}y^2 \sum_{i<j} \alpha_i\alpha_j \plus{} \ldots$.

Also, $ \int^1_0\int^1_0 x^{n\minus{}k}y^kdxdy \equal{} \frac{1}{(n\minus{}k\plus{}1)(k\plus{}1)}$

So the answer is $ \sum^n_{j\equal{}0} \frac{1}{(n\minus{}j\plus{}1)(j\plus{}1)}\sum_{1\leq i_1<\cdots<i_j \leq n} \alpha_{i_1}\cdots\alpha_{i_j}$.  I don't know if that simplifies at all.


Is there a way to make the parenthese look better?

$ (\frac{1}{2})$

Use \left( and \right) to adjust the sizes automatically to match the height of the enclosed formula: $ \left(\frac 12\right)$.


$f(x) = ax^2 + bx+c$ satisfy $|f(x)|\le1$ for all $x$ in $[0,1]$. Find the maximal value of $|a| + |b| + |c|$

I'm formulating a proof.  That seems to be leaning towards this answer.
[hide="Answer"]17[/hide]
Could you tell me if this is correct?

Yes it is correct.The polynomials $p(x)=8x^2-8x+1$, or $p(x)=-8x^2+8x-1$ has this property.Here the proof goes:
[hide]Suppose that there exist $q(x)=ax^2+bx+c$ with $|a|+|b|+|c|>17$ we may assume that $a>0$(if not then just take the polynomial -q(x)), then it is easy to see that $|c|\leq 1$
Also if $|a|>17$ then the polynomial $r(x)=p(x)-\frac{8}{a}\cdot q(x)$ is a polynomial of degree not more than 1.
But $r(0)>0,r(1/2)<0,r(1)>0$ contradiction since $r$ has 2 distinct zeros.So $|a|<8$
Then $|b|>8$.Also $b=-|b|$.Next $-1\leq p(1/2)=a/4+b/2+c\leq 3+b/2$ that gives $|b|\leq 8$ contradiction so there's no such polynomial $q$.[/hide]

Let  $f\left( x \right) = ax^2  + bx + c$ be a function, and we have:
$\left| {f\left( x \right)} \right| \le 1,\forall x \in \left[ {0,1} \right] \Leftrightarrow f\left( {\left[ {0,1} \right]} \right) \subseteq \left[ { - 1,1} \right]$
In this case, how we find $\max \left( {\left| a \right| + \left| b \right| + \left| c \right|} \right)$ ???

(I see that $\max \left( {\left| a \right| + \left| b \right| + \left| c \right|} \right)=17$)


If $2(4^{x})+6^{x}=9^{x}$ and $x=\log_{2/3}a$, find the numerical value of $a$.

divide both sides by 9^x to get:

2(4/9)^x+(2/3)^x-1=0...putting in a, we have

2a^2+a-1=0, and a>0...so

$a=\frac{1+\sqrt{3}}{2}$


inequalities
trigonometry
analytic geometry
graphing lines
slope

Prove that
$ cos{1}\plus{}(sin{1}\minus{}1)(\sqrt{2}\plus{}1)>0$

$ \cos{1} \plus{} (\sin{1} \minus{} 1)(\sqrt {2} \plus{} 1) > 0$

$ \Longleftrightarrow \frac {\sin 1 \minus{} 1}{\cos 1} > 1 \minus{} \sqrt {2}$

The left side means the slope of the line passing through some two points and the right side means $ \minus{}\tan 22.5^\circ$.


induction
graph theory
combinatorics unsolved
combinatorics

Let $G$ be a graph such that if you remove any $k-1$ vertices the graph will still connected. Prove that for any 2 vertices in $G$ there exist $k$ disjoint paths between them.

I proved a specific case for the 2-connected graph by induction on the distance between two vertics but i am having problems trying to generalize it.

I found this theorem with three proofs in book "Graph Theory" by R. Diestel:

Let $G=(V, E)$ be a graph and $A, B \subseteq V$. Then the minimum number of vertices separating $A$  from $B$ in $G$ is equal to the maximum number of disjont $A-B$ paths in $G$.

And as your graph is $k$-connected the result follows. So I think you can google this theorem - as you wrote in source this is Menger's Theorem.

I have the diestel book.. I will look it there.. thanks ;)


I remember taking a peek this problem:

"The spinner has only 3 and 8 on it. What is the largest possible value that can not be obtained?"

I remember the solution being 11 or 13 and it said that it can be solved by $ mn \minus{} m \minus{} n$ or something like that.

I didn't get it. Can someone explain?

Maybe the problem was "the largest possible value which [u]can't[/u] be obtained, with as many spins as you want. Then you would use mn-n-m to get 13. You can't get 13, but you can get any number higher (for example 14=8+3+3).

Isn't that supposed to be "cannot be obtained"?

Modular arithmetic is used kinda often here.
[hide]
Anyways, the $ mn\minus{}m\minus{}n$ is the answer for this, where $ m\equal{}3$ and $ n\equal{}8$.  This is known as the Chicken McNugget Theorem.

We know that all numbers congruent to 0 mod 3 greater than or equal to 3 can be obtained with just 3s.  If we have one 8, and then the rest of the numbers being 3s, then all numbers congruent to 2 mod 3 greater than or equal to 8 can be obtained.  Then, by the same process, all numbers greater than or equal to 16 that are congruent to 1 mod 3 can be obtained.  So subtract 3 from 16 to get [b]13[/b], which is the answer given by the formula above.
[/hide]

[quote="alanchou"]Isn't that supposed to be "cannot be obtained"?

Modular arithmetic is used kinda often here.
[hide]
Anyways, the $ mn \minus{} m \minus{} n$ is the answer for this, where $ m \equal{} 3$ and $ n \equal{} 8$.  This is known as the Chicken McNugget Theorem.

We know that all numbers congruent to 0 mod 3 greater than or equal to 3 can be obtained with just 3s.  If we have one 8, and then the rest of the numbers being 3s, then all numbers congruent to 2 mod 3 greater than or equal to 8 can be obtained.  Then, by the same process, all numbers greater than or equal to 16 that are congruent to 1 mod 3 can be obtained.  So subtract 3 from 16 to get [b]13[/b], which is the answer given by the formula above.
[/hide][/quote]
@lokito and alanchou: Sorry, my memory wasn't clear. My apology.

Oh, I see. But Chicken McNugget theorem.. :wink:


OK, I know when we do AIMEs for practice, we wonder what is the relative difficulty level of the one we are taking. So rank your top 3 hardest AIMEs in order. It would help me a lot. :|
1. 1994
2. 1995
3. 1999

Also, comment on other people's lists if you want.

I've only taken 83, 84, 85, 86, 87, 89, 2000A. Here's the order in which I'd rank them, hardest to easiest.

1. 2000A
2. 85
3. 87
4. 83, 84, 86, 89: all very easy.

1. 1994
2. 1992
3. 199-something?

I only remember those two being particular difficult for me... that was a while ago though. I can say 1983-1990 are pretty much the easiest, though...

Anything from 1993-1997 were all fairly hard...my average on those is about 3 less than the others.

Hmm...

 1st place: 1994, 1999, 1993

On these two AIME's I solved 12 problems

2nd place: 1993, 1996, 1997, 1998, 2003A

On these I solved 13 problems total, by the time i figure out one of the problems left I ran out of  time.

So yeah, difficulty of an AIME does not affect me very much

Ironically my highest score out of these is 1999(no careless mistakes!), the one I considered to be one of the hardest. So really careless mistakes control how well you do on AIME.

beta I don't think your personal story of what happened to you in terms of careless mistakes affects the actual difficulty of the contest....

1. 1994

Can't think of another one that really sticks out.

My philosophy is that you should train, in part, for problems more difficult than the ones you will encounter.  With that in mind, the bunch of Mock AIMEs I wrote last year were very difficult. I tried to make them harder than the 1994 AIME (hardest real one in my opinion.)

Out of curiousity, what was the floor on the 1994? I'm guessing 5 or 6.

There was no such thing as floor until 2002, partly because the AMC 10 didn't exist until 2001? (Maybe 2000)

Ah. So it was determined by AMC12+AIME index?

You mean AHSME. AMC 12 and AMC 10 replaced the AHSME.

I'm confused. I've seen a number of problems from before 2001 posted on here that were titled AMC12 so-and-so. Did the poster just put the wrong name?

The name was changed in 2000, according to [url=http://www.unl.edu/amc/e-exams/e6-amc12/factoids12.html]their site[/url].

Now that people are preparing for the 2006 AIME I and AIME II, it would be interesting to hear more opinions about which previous AIMEs are most challenging, and which are most easy. If I remember correctly (please correct me if I'm wrong), last year's AIME I (that is, the 2005 AIME I) was quite approachable to test-takers who were familiar with the problem-solving literature. I remember that when I page through it as a coach it seemed to me that many of the problem types were familiar.

Personally, my favorite AIME is 1994. (Other than the 1980's, those are too easy :P). 1991 is really fun too, along with 1990 and 1992. 1993 is a bit harder than the others though, and 1995 is the trickiest for me. In fact, it seems that all the 90's AIMEs have good problems... maybe because of all the algebra problems and only a few geometry ones (with the exception of 1995, which is the one I think is hardest :lol:)

Ironically, everyone else thinks that 1994 was hardest. :huh: I haven't taken any of the 2000-2005 AIMEs recently but I think I got like 4 or 5 on 2001 and 2002 a few months ago. :blush:

If there's a bunch of geometry problems on AIME 2006, I'll do really bad. :( :(


Okay, this is the player list. Order matters:

cf249
mathelete
chenshi
Wickedestjr
ernie
aleph0

Each player has 3 grenades in their arsenal. There is no concept of turns. No consecutive turns. When you go, you have five options. Do only one of them.

1. Choose another player, and subtract points. The number of points subtracted is the post number.

2. Same as above, but add points instead.

3. Throw a grenade. Convert the post number to binary, then add up the digits. The grenade will go off that many posts later. If it does not go off in 4 days, the grenade is defused.

4. Bump the grenade to the player below you on the list. If you are at the bottom, it will go to the top.

5. Pick up a defused grenade. It goes into your arsenal.

Everybody starts off with 0 points, and they can have a negative number of points. After 75 posts, the game ends. When the grenade goes off, whoever it is by falls for the rest of the game. If a person brings another person's points back to 0, then that person earns a mushroom. If a person doubles a positive number of points, that person earns a star. These are the sixth and seventh options. A mushroom gives yourself points. When a star is used on post x, the person is invincible till post 2x. An invincible person automatically bumps a grenade without posting.

No double posting.

cf249: 0, 3 grenades
mathelete: 0, 2 grenades
chenshi: 0, 3 grenades, GRENADE 5 BEEPING
Wickedestjr: 0, 3 grenades
ernie: 0, 3 grenades
aleph0: 0, 3 grenades

Throw grenade at chenhsi. It will go off at post 5, so I will call it grenade 5.

cf249: 0, 2 grenades
mathelete: 0, 2 grenades
chenshi: 0, 3 grenades, GRENADE 5 BEEPING
Wickedestjr: 0, 3 grenades
ernie: 0, 3 grenades, Grenade 5b beeping
aleph0: 0, 3 grenades

Throw grenade at ernie.  It will also go off on post 5 (4=100 in binary), so I will call it Grenade 5b.

cf249: 5, 2 grenades
mathelete: 0, 2 grenades
Wickedestjr: 0, 3 grenades
aleph0: 0, 3 grenades

Nice job helping with the double kill. For that you get 5 points.

wtf

I didn't even realize I was in this game yet.

And look at my comments in the other thread,

cf249: 12, 2 grenades
mathelete: 0, 2 grenades
Wickedestjr: 0, 3 grenades
aleph0: 0, 3 grenades

cf249: 12, 2 grenades
mathelete: 0, 1 grenade
Wickedestjr: 0, 3 grenades, GRENADE 9 BEEPING
aleph0: 0, 3 grenades

Throw grenade 9 at wicked. Either wickedestjr or aleph0 will be killed.

cf249: 21, 2 grenades
mathelete: 0, 1 grenade
Wickedestjr: PWNT
aleph0: 0, 3 grenades

cf249: 21, 2 grenades
mathelete: 0, 1 grenade
aleph0: 10, 3 grenades

give aleph 10 points

cf249: 21, 2 grenades
mathelete: 11, 1 grenade
aleph0: 10, 3 grenades

Ernie and chenshi tied for 5th place
Wickedestjr in 4th place

Let us finish this

cf249: 21, 2 grenades
mathelete: 11, 1 grenade
aleph0: -2, 3 grenades

-12 from aleph

cf249: 21, 2 grenades
mathelete: 24, 1 grenade
aleph0: -2, 3 grenades

cf249: 21, 2 grenades
mathelete: 24, 1 grenade
aleph0: -16, 3 grenades

Take away 14 points from aleph0.

Hey! I want to know why I didn't get notified that it restarted. Don't you think it's unfair that someone can lose before they even knew that a game existed?

cf249: 21, 2 grenades , GRENADE 17
mathelete: 24, 1 grenade
aleph0: -16, 2 grenades

Ernie and chenshi tied for 5th place
Wickedestjr in 4th place
cf249 in 3rd place

mathelete: 24, 1 grenade
aleph0: 1, 2 grenades

give aleph 17 points. Since grenades can be stalled in one-on-one, we will convert them to points. So:

mathelete: 25
aleph0: 3

Let us speed this up by ending it at post 32

mathelete:25
aleph0:22

give 19 points to aleph

chenhsi, I put it in the previous thread. You were too late.

BTW, you can pass your turn if you want. just say pass

[quote="aleph0"]mathelete:45
aleph0:22

How does someone win?[/quote]
By post 36, get more points.

BTW chenhsi took up a post, so this is really post 20. I pass

mathelete:24, star
aleph0:44

Since this is post 22 (chenhsi took up a post by being mad), give aleph 22 points. I gave aleph the lead to earn a star. I plan to use it later.

grr...

We don't even get time to be notified that the game restarted? You don't even ask if we still want to play?

And wth are you adding new rules in the middle of the game?

mathelete:48, star
aleph0:44, star

I lose...

chenshi, you wasted another post. I ended up redoing everything because people did not follow directions. This is really post 25. I use the star. Since it would normally last the rest of the game, I will make it last till post 32 to be nice. Now you can only give me points till then.

mathelete:48, invincible
aleph0:44, star(not used yet)

That's not my point. I have no objections with you restarting the game. My points is that you should have notified us before the second game started, and you should have allowed more players. And I dislike the fact that you ended a game without asking the players. What if we wanted to continue playing the game, even if we aren't using your rules?

Since I earned the star at 22, and aleph earned it at 24, we will convert them to points, so I win by 2!

1st-mathelete
2nd-aleph
3rd-cf249
4th-wicked
5th-chenhsi and ernie

I will post a new topic. Chenhsi, you are in.


There are positive integers $ k$, $ n$, and $ m$ such that $ \frac{19}{20} < \frac{1}{k}\plus{}\frac{1}{n}\plus{}\frac{1}{m} < 1$. What is the smallest possible value of $ k\plus{}n\plus{}m$?

I brute forced my way though this problem during practice, but does anyone have a solution?
 :lol:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1742539080&t=77789]Here[/url]'s some discussion, but I didn't find the explanations too satisfying.

[hide="My Solution"]
WLOG, let $ k\le n\le m$.
$ k\equal{}1$ will make this at least $ 1$, which doesn't work.
$ k\equal{}2$ means that $ m\ge n\ge3$, since if either of them were $ 2$ it would be at least $ 1$. If $ n\equal{}3$, then some testing gives the smallest possible value of $ m$ that works as $ 7$, so $ (2,3,7)$ works. If $ n\equal{}4$, then $ m\equal{}4$ would make this equal to $ 1$ and $ m\ge4$ would make this less than or equal to $ \frac{19}{20}$, both of which don't work. $ n\ge5$ would make this expression less than $ \frac{19}{20}$.
$ k\equal{}3$ gives the maximum at $ (3,3,3)$, but this equals $ 1$ and doesn't work. If we change it to $ (3,3,4)$, it's a bit less than $ \frac{19}{20}$, and any more increasing will make this value even smaller.
$ k\ge4$ makes the expression less than or equal to $ \frac34$.
We conclude that $ 2\plus{}3\plus{}7\equal{}\boxed{12}$ gives the minimum value.
[/hide]


The quesiton below is from '94 State competition, Team Round.

[img]http://www.geocities.com/danb80/quest.6.jpg[/img]

Can anyone help me figure out how to solve it?

[hide]the answer is 91[/hide]

Thanks!

[hide="hint"]inscribe the circles inside a rhombus. You should be able to find the length of one of the diagonals as well as one of the sides. Then find the other diagonal using rhombus properties and then the area of the rhombus can be found. [/hide]

Not [i]exactly[/i] sure about this but...

[hide]
Let's assume that the circles have radius 1.

Now, draw an equilateral triangle with side length 2 from the 3 circles as shown.  It should be clear that the whole plane can be tiled with that triangle.  The area of this triangle would be $\sqrt{3}$.  Now find the area of the sector bound by the sides of the triangle in one of the circles: $\frac{\pi}{6}$.  Since there are 3 such sectors multiply that by 3 to get $\frac{\pi}{2}$.  Divide that by $\sqrt{3}$ to get $\frac{\pi\sqrt{3}}{6}\approx0.90689968211710892529703912882108\approx91\%$
[/hide]

Thanks!

I thought of the rhombus solution, but it doesn't account for the other circles around.

The triangle method seems to work!

[hide]
Do the triangle thingy like ccy said.

You can say that the circles have a radius of $1$
Thus, the area of the triangle is $\frac{4\sqrt{3}}{4}=\sqrt{3}$
The amount of the circles that is inside the triangles is $\frac 12$ of a circle which is $\frac{\pi}{2}$
Thus, for every $\sqrt{3}$ triangle, $\frac{\pi}{2}$ is covered.
The answer is thus
$\frac{\frac{\pi}{2}}{\sqrt{3}}\approx.906\approx91\%$[/hide]

This problem has already been posted.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=57395&highlight=


The graph of $ y\equal{}ax^2\plus{}bx\plus{}c$ passes through points $ (0,5)$, $ (1,10)$, and $ (2,19)$. Find $ a\plus{}b\plus{}c$.

We're given the answer: let $ f(x)\equal{}ax^2\plus{}bx\plus{}c$; $ f(1)\equal{}a\plus{}b\plus{}c\equal{}\boxed{10}$.

[quote="GameBot"]The graph of $ y \equal{} ax^2 \plus{} bx \plus{} c$ passes through points $ (0,5)$, $ (1,10)$, and $ (2,19)$. Find $ a \plus{} b \plus{} c$.[/quote]

Second point: $ 10\equal{}a\plus{}b\plus{}c$.


University of Chicago
Stanford
college
MIT

Can anyone give any information about Northwestern U ? How good are the academics for math/physics and also the ISP programme? How does it compare with UChicago ?

I think I can help you out.  I applied to and got into both Northwestern and ISP, however, I'm actually going to be a freshman this fall at Caltech, so I can't say I have a 100% accurate view.  NU is a very well respected research university, with excellent facilities, especially in nanotechnology and bio-engineering related stuff.  Their chemistry department seems to be pretty amazing, and ISP kids usually either double major in chemistry, or drop out altogether of ISP and do chemistry.  Also, physics there seems to be pretty respected (one physics major I met was a goldwater scholar and going to Stanford for grad), though maybe not as much as chemistry.  I don't know about math (but I imagine that it's pretty decent too, though not UChicago)

As for ISP itself, it'll give you a very solid background in the sciences, though moreso in math and physics than biology.  It's a 3 year program that basically covers the same core class requirements of Caltech and MIT, and then goes a little beyond.  It's almost as hard as those respective classes at MIT and Caltech, though no Techer will ever admit it.  It's really easy to double major with ISP in math, chem, bio, physics, and materials science, and possible (though not as common) in other scientific or engineering majors.  Also, ISP gives more respect and status in terms of getting attention for research at NU, and all ISP classes are taught by professors in classes of 20 or less people.

The kids I met at NU were all fairly bright and talented one way or another.  The ISP kids I met were basically kids who were certainly smart enough to get into say MIT or Caltech but didn't due to the randomness of admissions.  ISP seniors were mostly going to grad school, and Stanford seemed to be the popular choice among the ones I met.  The acceptance rate for ISP applicants was 30% (and they had to get into NU before being considered for ISP).  Also, some ISP kids do become Goldwater and Gates scholars and whatnot.

NU has a huge rivalry with UChicago, and I'd say UChicago probably has a very slight edge on NU for science, though ISP seems to bridge that gap.  UChicago math is pretty incredible.  However, I only visited UChicago once, so what I know about them is mostly from rumors.

Sorry, it got to be a bit long.  Hope this helps?

Thanks a lot. I was deciding whether to add NU to my already long list.
I was also wondering, why is NU not that well known (on this forum for example)? UChicago also seems more popular than NU.

According to the website, it is need-blind for citizens and is a member of the common application. So, applying would make [i]some[/i] sense
But won't the high school GPA be very important there? We have low GPAs in high school. Is there any need-blind college where the high school GPA is [i]relatively[/i] less important especially when the gpa is from a foreign country?(in my case, India) The rest of my application (SAT, recommendations, etc) is alright, though not outstanding.

I'm sure plenty of the people here have heard of NU, it's just not talked about as much because most people here want to be math majors, and UChicago is considered a "top 5 math school", and thus many people here would probably look at UC for math, but not NU.  NU however seems to be a more well rounded school, as it has an engineering school, and has plenty of special programs for applied math and science (like ISP, MMSS, etc.), that are especially valuable for later careers in the financial sector or MBA programs.  Unless you want to go hard core theoretical math, I think you really can't go wrong choosing between NU and UC.

I would think all need-blind schools consider GPA very important, as most of them are rigorous and need to see that you can do the work.  That said, schools that are members of the common app are supposed to be committed to "holistic admissions", and look at your application in the context of your background, ie, they would see that schools in your country give lower GPA's, and would take that into account when considering your application.  That said, it is very competitive for international students to gain admissions to most needblind schools, but I can't give you any specifics, as there are so many different schools out there with slightly different admissions policies.

Actually, I wanted to major in pure mathematics. So U.Chicago might be a better place for me than Northwestern. But I lack outstanding achievements like IMO (didn't do well at the selection camp) and my GPA is low. Hence, getting into a high ranked college might be difficult.
However, I am [i]not[/i] an international student (although my school education has been from India). So getting into a need-blind college is what I have to worry about.


Other than using web.archive.org, can anybody tell me where the most complete compilation of old national competition results is? Thanks.

Please...I would really like these results. Thanks.

Maybe he wants to build a time machine, go back in time, make the winners really sick before the competition and make his state win..... :lol:  :P


calculus
integration
inequalities
function
real analysis
real analysis solved

The function $f(x)$ satisfies $f'(x)\geqq0$  at  $-1\leqq x\leqq 1$.
Prove the following inequality.

\[\int^1_{-1}\frac{x}{\sqrt{1+x^2}}f(x)dx\geqq0\]

Sorry for my misstyping.

I have just edited.

Best regards.

kunny

It's simply integration by parts. The LHS is $\int^1_{-1}(\sqrt{x^2+1})'f(x)dx=\sqrt 2\cdot(f(1)-f(-1))-\int^1_{-1}\sqrt{x^2+1}f'(x)dx$. We now use the fact that $f'$ is non-negative and the fact that $\sqrt{x^2+1}\le\sqrt 2$ to prove that $\int^1_{-1}\sqrt{x^2+1}f'(x)dx\le \sqrt 2\cdot\int^1_{-1}f'(x)dx=\sqrt 2\cdot(f(1)-f(-1))$, Q.E.D.

You are right, grobber.

This problem is a piece of cake for you,isn't it?

kunny


Given a triangle ABC and a point P on AB with AP>PB, Show how to construct a point Q on AC such that the triangle APQ will have one-half the area of triangle ABC.

I have gotten to the point where I drew a perpendicular line from H on AB that goes to point C, then a parallel line above this line will intersect AC at Q and AB at F. I am now stuck trying to determine how to find the distance between H and F which would enable drawing a parallel line to find Q. Any ideas??

$()$ denotes the area:

$(ABC) = \frac{1}{2}AB \cdot AC \cdot \sin{ \widehat{A}}$

$(APQ) = \frac{1}{2}AP \cdot AQ \cdot \sin{ \widehat{A}}$

$(ABC) = 2(APQ)\Rightarrow$

$AB \cdot AC = 2AP \cdot AQ \Rightarrow$

$\frac{AQ}{AC} = \frac{AB}{2AP}$

one way to find the point $Q$ is:

[list][b]-[/b] find a point $F$ on the ray $(AB$ such that $AF=2AP$

 (since $AP>PB$ you'll have $2AP>AB$, so $F$ is out of the segment $AB$)[/list]
[list][b]-[/b] connect $F$ and $C$[/list]
[list][b]-[/b] bring a parallel from $B$ to $FC$. This line will intersect the segment $AC$ at the point $Q$[/list]


analytic geometry
graphing lines
slope
geometry

A line with slope 6 bisects the area of the unit square with vertices (0,0); (0,1); (1,1); and (1,0). What is the y-intercept of this line?

[quote="4everwise"]... bisects the area of the unit square with vertices (0,0); (0,1); (1,1); and (1,0) ...[/quote]
By this do you mean that the line with slope six runs through the square with given vertices  :?

[hide]Since the line bisects the area of a square, it should pass throught center, which is $(\frac{1}{2},\frac{1}{2}$.

$y=mx+b$

$\frac{1}{2}=3+b$

$b=-2\frac{1}{2}$[/hide]


to continue the over-personal questions that would allow a stalker to find out more about you than should be known, how much does your car insurance cost per month. i have to pay $155 a month for minimum insurance. and i only drive a 96 neon. i think i'm getting ripped off.

I'm a male 17 year old student and I been in an accident...

 :(

I have not got the chance to drive a car... ... :(

Can't drive, don't really want to.  Well, I suppose that I could get a license, but I don't feel like paying for a car and insurance.

Haha...still waiting on that 06 model of the Mercedes SL65 AMG. *hop*

[quote="miraculouspostmaster"]I drive a 94 Land Cruiser and am paying less than you; yeah, I think you're getting ripped off.
BTW, have you submitted your grades to the insurance company?  Some will give you a discount for that sort of thing.  (I also had driver's ed, so mine is probably lower than normal).[/quote]

Do you currently live in the US. Here (according to some european friend, and my own experience) insurance is, first, in general, more expensive, second, a lot more expensive if you are younger than 25! (and so am I!)

Best,


real analysis
function
real analysis unsolved

Prove that if $ E \subset \mathbb{R}$ is a bounded set of positive Lebesgue measure, then for every $ u < 1/2$, a point $ x\equal{}x(u)$ can be found so that \[ |(x\minus{}h,x\plus{}h) \cap E| \geq uh\] and \[ |(x\minus{}h,x\plus{}h) \cap (\mathbb{R} \setminus E)| \geq uh\] for all sufficiently small positive values of $ h$.

[i]K. I. Koljada[/i]

for the first ques, i wonder if it's true. Consider the following measure supported on [0,1]. $ A_{1}\equal{}[\frac{1}{8},\frac{2}{8}]\bigcup [\frac{6}{8},\frac{7}{8}]$. $ A_{1}$ seperate [0,1] into three parts. $ B_{1}\equal{}[\frac{1}{64},\frac{2}{64}]\bigcup [\frac{6}{64},\frac{7}{64}]$, that is, analog to $ A_{1}$ respect to $ [0,\frac{1}{8}]$.Make $ B_{2}$ in the same way from $ [\frac{2}{8},\frac{6}{8}]$, and B3…… Generally, each step, add a set made in the above way from every section that has not been contained in E into E, cut a set (as we do to [0,1] in the first step) from every section that has already in E. It's obvious that the measure conserves to be $ \frac{1}{4}$. But it seems for every x and h, if (x-h,x+h) $ \in E$ then $ |(x\minus{}h,x\plus{}h)\cap E|\equal{}\frac{1}{4}h$

as to the second ques. Since E is bounded then every x far away enough from E satisfy $ |(x\minus{}h,x\plus{}h)\cap (\mathbb R  \setminus E)|\equal{}2h$

Sory for the wrong answer. The set I made was not definite. Maybe you can try Rusin theory.

Solution:
We will prove problem for every $u<1/4$, let $u=1/4-\varepsilon$ for $\varepsilon>0$. First let's find interval $I^1:\frac{|I\cap E|}{|I|}=\frac{|I\cap (\mathbb{R}\setminus E)|}{|I|}=\frac{1}{2}$. After Lebesgue's density theorem we get that there exists point $a:\exists lim_{h\rightarrow 0}\frac{|(a-h,a+h)\cap E|}{2h}=1$, so there exists $h:\frac{|(a-h,a+h)\cap E|}{2h}>\frac{1}{2}$, E is bounded set, so there exists point $b:\frac{|(a-h,a+h)\cap E|}{2h}=0<\frac{1}{2}$, denote function $\phi:[a,b]\rightarrow \mathbb{R}$, where $\phi(x)=\frac{|(x-h,x+h)\cap E|}{2h}$, $\phi\in C([a,b])$, so there exist point $c\in (a,b):\frac{|(c-h,c+h)\cap E|}{2h}=\frac{1}{2}$, so $I^1=(c-h,c+h)$. Let intervals $I_1,I_2:|I_1|=|I_2|=\frac{|I^1|}{2},I_1\sqcup I_2=I^1$, easy to see that $min(\frac{I_1\cap E}{|I_1|},\frac{I_2\cap E}{|I_2|})\leq \frac{1}{2}\leq max(\frac{I_1\cap E}{|I_1|},\frac{I_2\cap E}{|I_2|})$, so there exist interval $I^2:\frac{|I^2\cap E|}{|I^2|}=\frac{1}{2}-\delta,|\delta| \leq2\varepsilon,[I^2]\subset  I^1,|I^2|=|I_1|=|I_2|=\frac{|I^1|}{2}$. Like the same construct intervals $I^2,I^3,...$, where $[I_{n+1}]\subset  I_n, |I_{n+1}|=\frac{|I_n|}{2}\ \forall n$, let $x(u)=\cap_{i=1}^{\infty}I^{i}$, $x(u)\in I^1$, so there exist $\tau >0:O_{\tau}(x(u))\subset I^1$, for every $h<\tau$ there exist $n_h\in N:I^{n_h}\subset O_h(x(u)),|I^{n_h}|\geq \frac{h}{2}$, so $|(x(u)-h,x(u)+h)\cap E|\geq |I^{n_h}\cap E|\geq (\frac{1}{2}-2\varepsilon)|I^{n_h}|\geq (\frac{1}{4}-\varepsilon)h=uh$, like the same $|(x(u)-h,x(u)+h)\cap (\mathbb{R}\setminus E)|\geq uh$. done


superior algebra
superior algebra solved

Show that the cyclotomic fiedl E8=Q(i, √2).

Thank you very much!

It is easy to prove that the $8$-th cyclotomic field is contained in $\mathbb Q(i,\sqrt 2)$. Just write down the $8$-th roots of unity to see this.
We have then $\mathbb Q\subseteq E_8\subseteq \mathbb Q(i,\sqrt 2)$. We have $[E_8: \mathbb Q]=\varphi(8)=4$ and $[\mathbb Q(i,\sqrt 2): \mathbb Q]=[\mathbb Q(i,\sqrt2): \mathbb Q(i)]\cdot[\mathbb Q(i): \mathbb Q]=2\cdot 2=4$. It follows $E_8=\mathbb Q(i,\sqrt 2)$.


Jerry wins the race, but he gets no trophy.

How is this possible?

[hide]Jerry is a horse. His rider gets the trophy[/hide]

[hide] He gets a medal or a cash prize or a certificate....or this is a horse race and jerry is the horse...i dunno! :D [/hide]


There are mountains of things to memerize in biology. It is not like math or physics, in which you can solve a lot of problems by talent. No hints of studying it, just memerize and do exercises.

[quote="furious"]just thought i'd get this started.
i'm in ap bio. who feels ap has too much memorization? i know i do. hope everyone is ready for the ap exam or will be on exam day.[/quote]

You just described why I lost interest in bio.

However, I do think computers are going a long way to bringing mathematics and rigor into biology, not to mention the vast amounts of money in biotech drawing better and better people into the field.  I know a handful of olympiad types who are now 'computational biologists', melding math, cs, and bio much the same way string theorists mix math and physics.

Exactly why I don't like biology! It's way too much memorization (that's so unscientific, even history is a bit more flexible)!

I dont like biology at all (same reason)..........but I want to be a doctor. Not only is it boring to memorize everything, i feel its also not very interesting. I hope this forum could clear up some confusion.

All sciences have some amount of memorization required for study. Biology has a lot of this, not because it wants to bore you with its repetitiveness, but because it is an inherently structured science. It is also by far the most expansive science. Studying ecology and evolution can be extremely different from studying development and genetics, and most people find their biology niche (no pun intended) and then delve into that specific area. Others like the complete unification that biology presents because though all the different parts of biology can be vastly different to study, evolution and molecular chemistry and processes are prevalent motifs and unify the whole idea of life sciences very well.

AP Biology is a class meant to get you to learn all the basics of biology and also to teach how all the different parts are unified by evolution. When you look into the specific topics that biology can present, you will understand the great magnitude of this science. How amazing is it that the pGLO gene present in some bioluminescent cnidarians can be transferred into Arabidopsis thaliana plants to make them "glow"? How amazing is it that the fusion of two haploid nuclei at fertilization can carry all of the information necessary to build a functional, living organism that has such complex cognition as man? These are the phenomena of biology and some things that AP Biology will present to you.

Don't let AP Biology turn you off to the science. In college, biology (especially laboratory) becomes less memorization and a whole lot more chemistry and physics (especially in the submicroscopic fields of biology). If you choose to do zoology or ecology, however, you probably enjoy the memorization of systematics but the laboratory experiences in those fields are also amazing.

In conclusion, not only is biology an ever-growing frontier science, it is the science that is most applicable to everyday life and the science that growing amounts are learning to cherish for its widespread scope and intrigue.

Just wondering, I want to be doctor as well (not sure what kind) but then, should I have my major on biology?

I guess biology is a lot of memorizing but it's actually pretty fun once you get the concept. Also, my biology teacher actually taught me one useful thing this year and that is "look for the words you know."

Like if you have "sporophyte" and you know "phyte" means plant and "sporo" means spore, you can see that the word "sporophyte" is a plant that produces spores for the reproduction. :)

But do I have to take major as biology if I want to be doctor? Is it recommended that way?

i'm not sure if your major has to biology, but i would assume it would be in a doctor's best interest to major in some branch of biology or science.
anywho, in ap bio, i enjoyed the first half of the year when we were learning celluar processes like respiration and photosynethesis as well as genetics. but i hate the second half of the year because it is 100% memorization - memorize plants, memorize invertabrate, memorize the body systems, etc. o well.

you definitely do not have to be a biology major to become a doctor - even an english or music major can apply for med school provided they have the required courses... which are usually covered by the core requirements at top universities...

[quote="tetrahedr0n"]Unfortunately, a modern AP biology class is essentially a "memorize a book" class. This kind of class has eliminated any thoughts of a potential biology career for me.[/quote]
I couldn't agree more.

a science degree is not even required to become a medical doctor, though it would be quite helpful. pre-med is a very common major for aspiring physicians.

yes.... many people find that there is lot of stress on memomorisation in biology
i agree with what silverfalcon said... :)  if the teacher said it becomes easier if you know the terminology, therefore it depends a lot on the teachers who explain the concepts if a little effort is showwn towards the understanding of the terminology in the lower classes it is easy to catch what the teachres are talking in the higher classes especially in the chapters like systematics as addidasty  ;)


LaTeX
inequalities
inequalities proposed

Let be $ a,b,c>0$ such that $ abc\equal{}1$ . Show that :

$ \frac{(a^2\plus{}2b^2)(b^2\plus{}2c^2)(c^2\plus{}2a^2)}{a^2b\plus{}b^2c\plus{}c^2a}\ge 9$

Hello,  :D

$ {a^2b \plus{} b^2c \plus{} c^2a}\geq 3 \sqrt [3]{a^3 b^3 c^3} \equal{} 3 \implies \frac {(a^2 \plus{} 2b^2)(b^2 \plus{} 2c^2)(c^2 \plus{} 2a^2)}{a^2b \plus{} b^2c \plus{} c^2a}\geq \frac {(a^2 \plus{} 2b^2)(b^2 \plus{} 2c^2)(c^2 \plus{} 2a^2)}{3}$

Now it is enought to prove that $ (a^2 \plus{} 2b^2)(b^2 \plus{} 2c^2)(c^2 \plus{} 2a^2)\geq 27$

By AM-GM

$ a^2 \plus{} 2b^2 \equal{} a^2 \plus{} b^2 \plus{} b^2 \geq 3\sqrt [3]{a^2 b^4} ,$

 $ b^2 \plus{} 2c^2 \equal{} b^2 \plus{} c^2 \plus{} c^2 \geq 3\sqrt [3]{b^2 c^4} ,$

 $ c^2 \plus{} 2a^2 \equal{} c^2 \plus{} a^2 \plus{} a^2 \geq 3\sqrt [3]{c^2 a^4} ,$

Multiplying we have  $ {(a^2 \plus{} 2b^2)(b^2 \plus{} 2c^2)(c^2 \plus{} 2a^2)}\geq 27\sqrt [3]{a^6 b^6 c^6} \equal{} 27$ as desired.

By the way ,How to center LaTeX ?

Sorry: IGNORE THIS POST !

[quote="Jianxing113725"]But you are wrong in the first step... :maybe:[/quote]

Okay!..It's not easy  :lol: ..
I will try

[quote="enndb0x"]Can you tell me where exactly is my mistake.Is it typo,or it is wrong ?[/quote]

$ (a^2\plus{}2b^2)(b^2\plus{}2c^2)(c^2\plus{}2a^2) \ge 27$
But:
$ 9(a^2b\plus{}b^2\plus{}c^2) \ge 27$

i think the condition  $ abc\equal{}1$ is not correct... and i think the right condition should be $ abc\ge 1$ for  $ a,b,c>0$

and that could be proven easily using enndb0x's method with some small improvements

So... that's it :D:D

You're right . can you post your solution ?

What you actually say is that :

$ A>B$ and $ A>C \Rightarrow B>C$ (which can be false) , where $ A\equal{}\frac{(a^2\plus{}2b^2)(b^2\plus{}2c^2)(c^2\plus{}2a^2)}{3abc}$ , $ B\equal{}\frac{(a^2\plus{}2b^2)(b^2\plus{}2c^2)(c^2\plus{}2a^2)}{a^2b\plus{}b^2a\plus{}c^2a}$ and $ C\equal{}9$

[quote="SHP3ND1"]Well ... if the expression above is greater or equal than $ 27{a^{2}b^{2}c^{2}}$   and the expression under that is greater or equal than $ 3 abc$      ... then if we divide first with the second we will have :

$ \frac { 27a^{2}b^{2}c^{2}}{3abc}$  which is greater or equal with $ 9$ for $ abc\ge 1$

So... this is right... i can't see any mistake here  :P  :rotfl:[/quote]
You have a mistake my friend...
$ a^2b + b^2c + c^2a \geq 3abc$
But $ \frac {1}{a^2b + b^2c + c^2a} \leq \frac {1}{3abc}$
So you prove that an expresion is greater that $ 9$,but simultanously this expresion is also greater that what you have to prove..
Let me put another way:we have to prove that $ b \geq c$...If we find a number $ a$ so that $ a \geq b$ and $ a \geq c$ it doesnt mean that $ b \geq c$....

[quote="alex2008"][quote="Dimitris X"]
Let me put another way:we have to prove that $ b \geq c$...If we find a number $ a$ so that $ a \geq b$ and $ a \geq c$ it doesnt mean that $ b \geq c$....[/quote]

I guess you didn't see my post . I explained him the same thing :P[/quote]
Yes...I saw your post after my reply :P

[quote="alex2008"]Let be $ a,b,c > 0$ such that $ abc \equal{} 1$ . Show that :

$ \frac {(a^2 \plus{} 2b^2)(b^2 \plus{} 2c^2)(c^2 \plus{} 2a^2)}{a^2b \plus{} b^2c \plus{} c^2a}\ge 9$[/quote]
After expanding and homogenizing:
$ \sum_{cyc} 4a^4b^2\plus{}2a^2b^4\plus{}3a^2b^2c^2 \minus{} 9a^3b^2c \ge 0 \iff$
$ \left ( \sum_{cyc} 8a^4b^2\plus{}a^4c^2\plus{}a^2b^4\plus{}8a^2b^2c^2 \minus{} 18a^3b^2c \right ) \plus{} 2\left ( \sum_{cyc} a^2b^4\minus{}a^2b^2c^2 \right)\ge 0$ Which is true by Am-GM :)

Let $ a \equal{} \frac{x}{y}, b \equal{} \frac{y}{z}, c \equal{} \frac{z}{a}$ and substitute to the inequality, multiply numerator and denominator by $ (xyz)^4$. After expanding, the desired inequality becomes:

$ 3 \sum x^4y^4z^4 \plus{} 2\sum x^8y^2z^2 \plus{} 4 \sum x^6y^6 \geq 9 \sum x^6y^3z^3$ (where all summation is symmetric)

Now, from Schur's:
$ A^3\plus{}B^3\plus{}C^3\plus{}3ABC \geq \sum A^2B$, and let $ A \equal{} x^2y^2, B \equal{} y^2z^2, C \equal{} z^2x^2$ we have:

$ 3\sum x^6y^6 \plus{} 3\sum x^4y^4z^4 \geq  6\sum x^6y^4z^2$

So now we only need to prove:
$ 2\sum x^8y^2z^2 \plus{}  \sum x^6y^6 \plus{} 6 \sum x^6y^4z^2 \geq 9 \sum x^6y^3z^3$

But each term in the inequality above is true by Muirhead / AM-GM.

Your solutions are better than nothing , but there's no solution without brute force ?

[hide="P.S."]Actually hendrata01's substitution gave me an idea to corelate this inequality to another one which is well-known[/hide]

[quote="alex2008"]Your solutions are better than nothing , but there's no solution without brute force ?
[/quote]

ACtually my "brute force" isn't that bad, you just multiply 2 terms x 2 terms x 2 terms = 8 terms, and a lot of them are cyclic, so there's not that much algebra really. The reason the coefficients turned out that way is just because there's a "2" coefficient that gives the impression that it's a grungy algebra. It's not really

hendrata01 your method works in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=277327]this[/url] inequality ?

hey, so Hendrata still playing with this math problem..  :D  :D

(udah kerja kan ?? Masi sempet aja ya maen2 ke Mathlinks..  :P )

[quote="alex2008"]Let be $ a,b,c > 0$ such that $ abc \equal{} 1$ . Show that :

$ \frac {(a^2 \plus{} 2b^2)(b^2 \plus{} 2c^2)(c^2 \plus{} 2a^2)}{a^2b \plus{} b^2c \plus{} c^2a}\ge 9$[/quote]
with $ abc\equal{}1$ the problem isn't wrong and this is my slution

The inequality equivalent

$ (a^2 \plus{} 2b^2)(b^2 \plus{} 2c^2)(c^2 \plus{} 2a^2)\ge 9(a^2b \plus{} b^2c \plus{} c^2a)$

setting $ x\equal{}\frac{a}{c}$, $ y\equal{}\frac{b}{a}$, $ z\equal{}\frac{c}{b}$ $ \Rightarrow xyz\equal{}1$

assume $ x\equal{}min(x,y,z) \Rightarrow x\le 1$

Inequality became

$ (1\plus{}2x^2)(1\plus{}2y^2)(1\plus{}2z^2)\ge 9(x\plus{}y\plus{}z)$

setting $ F(x,y,z)\equal{}(1\plus{}2x^2)(1\plus{}2y^2)(1\plus{}2z^2) \minus{} 9(x\plus{}y\plus{}z)$

we have

$ F(x,y,z)\minus{}F(x,\sqrt{yz},\sqrt{yz})\equal{}(\sqrt{y}\minus{}\sqrt{z})^2[2(1\plus{}2x^2)(\sqrt{y}\plus{}\sqrt{z})\minus{}y]$

By Am-Gm $ \Rightarrow F(x,y,z)\ge F(x,\sqrt{yz},\sqrt{yz})$

we will prove $ F(x,\sqrt{yz},\sqrt{yz})\ge 0$

$ \Leftrightarrow F(1/t^2,t,t)\ge 0$

$ \Leftrightarrow (t\minus{}1)(4t^7\plus{}4t^6\plus{}8t^5\plus{}8t^4\plus{}17t^3\minus{}t^2\minus{}2t\minus{}2)\ge 0$ (right!!!)

DoNe.

[quote="alex2008"]Let be $ x,y,z > 0$ such that $ xyz \equal{} 1$ . Show that :

$ \frac {(x^2 \plus{} 2y^2)(y^2 \plus{} 2z^2)(z^2 \plus{} 2x^2)}{x^2y \plus{} y^2z \plus{} z^2x}\ge 9$[/quote]
Let $ x \equal{} \frac {a}{b}$... We need to prove:
$ (2a^4 \plus{} b^2c^2)(2b^4 \plus{} c^2a^2)(2c^4 \plus{} a^2b^2) \geq\ 9a^3b^3c^3(a^3 \plus{} b^3 \plus{} c^3)$
<=> $ 4\sum\ a^3b^3 \plus{} 9a^3b^3c^3 \geq\ 7a^3b^3c^3(a^3 \plus{} b^3 \plus{} c^3)$
Using Am-Gm and Schur ineq, we have: $ RHS \geq\ \frac {7}{2}a^2b^2c^2 \sum\ a^4(b^2 \plus{} c^2) \geq\ LHS$
(obviously trues by Am-Gm)
We have done  :)


Chan's bank account contains $ \$450$, which is $ 125\%$ more than
it contained at this time last year. How many dollars were in
Chan's bank account last year?

There is a typo in the question. The percentage in the question shows $ 125 $% while the correct percentage is $ 225$

No, there isn't. 125% [i]more[/i] is the same as 225% [i]of[/i].


How many integers between 300 and 500 have the sum of their
digits equal to 16?

If the hundreds digit is 3, then the other two digits add to 13. There are 6 ways there. If the hundreds digit is 4, then the other two digits add to 12. There are 7 ways. Added up, 13 ways.


I'm getting ready for GPML state, and I was wondering if anybody knows where I can find old GPML problems.  Also, does anyone know where to find ARML problems that are older than 2005

If your school has participated in GPML for a while, your math coach probably has copies of old tests.  I know that my school's math coach has quite a few...  Other than that, I don't think they're on the web anywhere, though if it's practice you want, going through old AMC 12s (available on this site) are about the same in difficulty, maybe a bit harder.  Also don't know about old ARML problems (there are probably books you can buy with past problems).  If it's just ARML-ish problems, though, and you've already done the ones since '05, there are quite a few mock ARMLs stored on AoPS... try the wiki, search the forums.

There's a neat thing called [url=http://lmgtfy.com/?q=gpml&l=1]Google[/url] :wink:


FTW
calculus
integration
counting
distinguishability
geometry
geometric transformation

Yay, my fav math tourney's coming up! (bcuz its uber long and they give ciphering prizes. also its PROOF. very hardcore :O)

So, who's going? I'm going JV, slackers ftw :D

Uh, like no one from Northview's going, cuz science bowl, debate, and science olympiad tourneys are on the same day.

Uh, I'm going to be daring and say GT is not my favorite tournament cuz the only time I participated, I got a perfect on the first round, did okay on ciphering, and made over 9000 careless mistakes on proof round.  The proof problems are more like "find the value and justify that it's the answer," so they're not very hardcore.  But that's just JV.  I dunno about varsity.  I guess it's just because I'm not 1337. :(

Question. I've heard rumors that there are no ciphering awards.

O RLY? :O

plz dun say YA RLY. :(

Ciphering is conducted towards team scores.  I don't know if there's a prize

Does anyone want to form an alternate team since I'll probably be the only person going from Norcross?

Hey, so when I was doing the 2005 proof paper I noticed that the solutions file was missing on the website. If it exists could someone tell me where I can get it?

If it doesn't, maybe we should just go over or outline the solution to each question here  :)
(Especially #5, I keep getting 0 for c) and my proof for b) is a little iffy)

[hide="Number 1"]

$ \binom{2n}{n}$ is the coefficient of $ x^n$ in $ (x \plus{} 1)^{2n}$

But

$ (x \plus{} 1)^{2n} \equal{} [(x \plus{} 1)^n]^2$

and so

$ \binom{2n}{n} \equal{} \sum^{n}_{j \equal{} 0}{\binom{n}{j} \binom{n}{n \minus{} j}} \equal{} \binom{n}{j}^2$

[/hide]

#5: This is pretty much "Have you heard of the Cantor Set, and do you understand some of the math behind it?" This is a rough outline of the solutions, sorry for not LaTeXing this.

[hide]a)
There are just as many numbers in the Cantor Set as in the interval (0,1)
Proof: 1 to 1 correspondence:
Every number in the Cantor Set when written in base 3 only contains 0's and 2's. So for every number in the Cantor Set, write the number in base-3, change all the 2's to 1's, and then erase the base-3 and write base-2 without actually changing the digits. This will yield a distinct number in the interval (0,1). You can do the same thing in reverse on any number in (0,1) to get a number in the Cantor Set. So, there is a 1:1 correspondence, and thus the same number of points in both the Cantor Set and (0,1).

b)
No, it is not possible.
Suppose that it was possible to have a 1:1 correspondence.
Let the number corresponding to m in the Cantor set be:
M: 0.d_{m,1}d_{m,2}d_{m,3}...d_{m,n}... in base-3
Also let d'{m,n} = 2-d_{m,n} (In other words, the compliment digit of d_{mn})
Remember that since numbers in the Cantor Set don't have 1's for digits in base-3, the compliment is a different digit than the original digit.

Consider the number X = 0.d'_{1,1}d'_{2,2}d'_{3,3}...d'_{n,n}....
The 1st digit after the decimal is d'_{1,1}, which is different from d_{1,1}, the 1st digit after the decimal of the number corresponding to 1. So, X doesn't correspond to 1.
The 2nd digit after the decimal is d'_{2,2}, which is different from d_{2,2}, the 2nd digit after the decimal of the number corresponding to 2. So, X doesn't correspond to 2.
And in general, The nth digit after the decimal is d'_{n,n}, which is different from d_{n,n}, the nth digit after the decimal of the number corresponding to n. So, X doesn't correspond to n.

So, this number X does not correspond to any natural number. Thus, our assumption that there was a 1:1 correspondence is false.

c) After each iteration, 1/3 of the length is removed, so 2/3 of the length remains.
So, the length after n iterations is (2/3)^n
Thus, the length of the Cantor Set is lim(n->infinity) (2/3)^n = 0.
Yes, the length is actually 0 despite the fact that some numbers, such as 1/4 and 10/13, are in the Cantor Set. I believe that numbers that are in (0,1) and are not in the Cantor are "dense" in the interval (0,1), but I am not sure that is the correct terminology.[/hide]

#2: I believe the simplest solution is casework on the number of quarters and dimes, and figure out the combination of nickels and dimes from this. Does anyone know of a better way?

#3: This solution uses calculus:
[hide]
If the original series converged, then its sum would be greater than the sum of the series $ \displaystyle \sum_{n = 1}^{\infty} \frac {1}{10n - 1} = \frac {1}{9} + \frac {1}{19} + \frac {1}{29} + \cdots + \frac {1}{90} + \frac {1}{99} + \frac {1}{109} + \cdots$ (sum of reciprocals of only the integers with 9 as the last digit)

If this series converged, then its sum would be greater than $ \frac {1}{9} + \frac {1}{10}\displaystyle \int^\infty_9 \frac {1}{x} \, dx$.

However, $ \int^\infty_9 \frac {1}{x} \, dx$ diverges, and thus, so does both $ \displaystyle \sum_{n = 1}^{\infty} \frac {1}{10n - 1}$, and the original series in the problem. [/hide]

#4: This is just Fermat's Little Theorem. I don't feel like proving this now, so instead, I will tell you this story:
[hide]
In a research base in Antarctica, $ p$ penguins (where $ p$ is a prime number) were held captive by researchers. One day when the researchers were particularly negligent, all of the penguins escaped. To celebrate their escape, these penguins decided to host a dinner party (as proper penguins would do.)

They all waddled over to the nearest tuxedo rental store, and found out that the store had $ a$ different colors of matching tuxedo and bow-tie packages. Since they were sick of being forced to conform to each other in the research base, they decided that each penguin could buy which ever color tuxedo and bow-tie that he or she wanted, as long as they wouldn't all be wearing the same color tuxedo.

After this, they went to a restaurant, and sat down randomly at a perfectly round table for $ p$. During the dinner conversation, one of the penguins posed a problem: "If the only distinguishable feature of the penguins was what color tuxedo they wore, and there was no way to distinguish rotations of the table, how many different dinner table arrangements could be made?"

"Well," Waddles said, "Since there are $ a$ tuxedo outfits and $ p$ of us, there are $ a^p$ different combinations of tuxedos that we could have bought."

Tux immediately interjected, "Don't forget that we agreed that we couldn't all wear the same color, so that leaves $ a^p - a$ possible combinations of tuxedos that we could have bought. But we still need to account for rotations."

Waddles replied, "That is easy. Since there are a prime number of us, each possible arrangement can either be rotated to yield $ p$ different arrangements, or only be rotated to yield the same arrangement over and over. But that can only happen if we all wore the same color tuxedo. So, there are $ \frac {a^p - a}{p}$ different table arrangements."

All of the penguins agreed that this was an interesting problem. After a few more hours of intellectual conversation, the penguins thanked their gracious host, and left the dinner party.
[/hide]
Anyway, now that I have finished the story, here is the rest of the proof.
[hide]
Clearly, $ \frac {a^p - a}{p}$, the number of different table arrangements, must be an integer.

So, $ a^p - a \equiv 0 \pmod{p} \Rightarrow a^p \equiv a \pmod{p}$

Since $ p \nmid a$, $ a \not\equiv 0 \pmod{p}$. Thus, $ a^{p - 1} \equiv 1 \pmod{p}$ as desired.

The story is similar to the beads and necklace argument, however, I felt that penguins were more interesting.
[/hide]

Your penguin proof of FLT is interesting!  Did you see the proof of FLT that AoPSer t0rajiru came up with?  Here's the link to that entry in his blog:

[url]http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=229442[/url]

Thanks, Santosh!

Yeah, for #2, I don't see any way you could've used generating functions without making a mess.

For #3, couldn't you just compare it to the harmonic sequence and prove that the harmonic sequence diverges by creating infinite groups of n>1/2? (yeah, that might take a little longer, though)

I liked your counting solution for #4  :D
I just ended up using $ (1\plus{}(a\minus{}1))^p \equiv 1^p \plus{} (k\minus{}1)^p \text{mod p}$

You don't need to prove that the harmonic series diverges; it's well known.

Yes, I think combinatorics and induction are two of the three well known elementary proofs of Fermat's little theorem.  Here is the third one.  We will let $ a$ be any integer not divisible by $ p$.

[hide]Notice that the sets
\[ \{ 1, 2, \dotsc, p-1 \} \]
and
\[ \{ a, 2a, \dotsc, (p-1) a \} \]
are congruent mod $ p$, since $ a$ has a multiplicative inverse.  Then their products are congruent mod $ p$; that is,
\[ (p-1)! = \prod_{k=1}^{p-1} ka = a^{p-1} \cdot (p-1)! . \]
Since $ (p-1)!$ has a multiplicative inverse, it follows that $ a^{p-1} \equiv 1 \pmod{p}$.[/hide]

Here is a group-theoretic approach.  If you read the background through the proof of Lagrange's theorem, it seems like kind of a lot, but know that it is a fundamental part of modern algebra, and if you ever take undergraduate algebra, you will learn all of it.  Besides, it is pretty, and what comes afterward is very natural; by now it is the easiest of these four proofs for me to remember.

[hide]
First we need to prove Lagrange's Theorem (well, sort of a special case), for which we need some background.  All group composition will be written multiplicatively, and the identity will be denoted $ e$.

[hide="Background"]
Let $ G$ be a group.  We call the the number of elements of $ G$ (the cardinality of the set of elements of $ G$) the [i]order[/i] of $ G$, and we denote it $ \text{ord}(G)$, or sometimes $ \lvert G \rvert$.

Let $ H$ be a subgroup of $ G$.  We say that two elements $ x,y$ of $ G$ are [i]left equivalent[/i] mod $ H$ if $ x^{-1}y \in H$.  (For those interested, right equivalence is defined by the relation $ xy^{-1} \in H$.)

[b]Proposition.[/b] Left equivalence mod $ H$ is an equivalence relation on $ G$.

[i]Proof.[/i] We need to establish that the relation is reflexive, symmetric, and transitive.  The relation is reflexive because $ x^{-1}x$ is the identity, which is in $ H$ by definition of "subgroup".  The relation is symmetric because if $ x^{-1}y \in H$, then so is $ (x^{-1}y)^{-1} = y^{-1}x$.  Finally, the relation is transitive because if $ x^{-1}y \in H$ and $ y^{-1}z \in H$, then
\[ x^{-1}z = (x^{-1}y)(y^{-1}z) \in H, \]
as $ H$ is closed under composition.  Thus left equivalence mod $ H$ is an equivalence relation on $ G$.  $ \blacksquare$

A [i]left coset[/i] mod $ H$ is a set of the form
\[ \{ x \in G \vert a^{-1}x \in H \} \]
for some $ a \in G$.  In other words, it is the set of all $ x\in G$ that are left equivalent (mod $ H$) to some fixed $ a \in G$.  More succinctly, a coset is an equivalence class under the relation of left equivalence mod $ H$.

The cardinality (number of elements) of the set of left cosets mod $ H$ is denoted $ (G: H)$.  (It is the same as the cardinality of the set of right cosets, but that's irrelevant here, since we're only going to talk about commutative groups anyway.)

Okay, that's all the background we need![/hide]

[b]Theorem (Lagrange).[/b] Let $ G$ be a group, and let $ H$ be a subset of $ G$.  Then
\[ \lvert G \rvert = (G : H) \cdot \lvert H \rvert . \]

[hide="Proof"]Since each element of $ G$ is an element of exactly one left coset mod $ H$, it is enough to show that every left coset mod $ H$ has the same size as $ H$.

To this end, let $ S$ be an arbitrary left coset mod $ H$, and let $ a$ be an element of $ S$.

Consider the mapping $ A: x \mapsto ax$ of $ H$ into $ S$.  This is indeed a mapping into $ S$ because for all $ x\in H$, $ a^{-1}(ax) = x \in H$.  It is injective, because if $ A(x) = A(y)$, then $ ax = ay$, so $ x=y$.

Furthermore, if $ b$ is an arbitrary element of $ S$, then $ a^{-1}b \in H$, and $ f(a^{-1}b) = a(a^{-1}b) = b$, so $ f$ is surjective.  Hence $ f$ is a bijection between $ H$ and $ S$.  Therefore all the cosets of $ S$ have the same size as $ H$, and we are done.  $ \blacksquare$[/hide]

Now, let $ G$ be the group of nonzero integers modulo $ p$, under multiplication.  Since $ G$ is finite, we must have $ a^{r} = a^s$ for some integers $ r>s$.  Then for some positive integer $ N$ (equal to $ r-s$), $ a^N = e$.  Then let $ n$ be the least integer such that $ a^n = e$.

The powers of $ a$ evidently form a subgroup of $ G$, and the order of this subgroup is evidently $ n$.  Since the order of $ G$ is $ p-1$, it follows from Lagrange's theorem that $ n$ divides $ p-1$.  It follows directly that $ a^{p-1} = e^{(p-1)/n} = e$, or, in the notation of modular arithmetic,
\[ a^{p-1} \equiv 1 \pmod{p} . \qquad \blacksquare \]
[/hide]

Now, are you ready for the slickest proof of Wilson's theorem?  Since the integers mod $ p$ form a commutative field, and all nonzero integers (mod $ p$) are zeros of the polynomial $ x^{p-1} - 1$, it follows that
\[ (x - 1)(x - 2) \dots p \bigl( x - (p-1) \bigr) \equiv x^{p-1} - 1  \pmod{p}. \]
Set $ x$ equal to zero.  The left-hand side of the equation evaluates to $ (-1)^{(p-1)}(p-1)! = (p-1)!$, and the right-hand side evaluates to $ -1$.  So
\[ (p-1)! \equiv -1 \pmod{p} . \]

Just as a general question, what can we assume that the graders will accept as being 'well known'?

Specifically, I was a little worried by my statement on a (practice) proof that:

It is well known that the Fermat point, which makes three 120 degree angles at the center of a triangle, minimizes the sum of the distances to the three sides.

In general, it is acceptable to cite any widely known result as "well-known" if it does not trivialize the problem at hand.

This is particularly true if the "well-known" results is indeed well known in an area of modern mathematics.  The divergence of the harmonic series is a fairly basic fact in the field of analysis, a vibrant and mainstream field of research these days.  The Fermat point is a fairly obscure corner of Euclidean geometry.

In any case, the Fermat point does [i]not[/i] minimize the sum of the distances from a point to the three sides of a triangle.  That would be the vertex opposite the longest side.

Oops  :blush:
I meant the distances to the three vertices...

But thanks for clarifying that for me, I think I get the general idea of what I can cite now.

[hide="non-calculus way for #3"]Note the series includes $ \sum_{k \equal{} 1}^{\infty}\sum_{n \equal{} 9\cdot 10^k}^{10^{k \plus{} 1} \minus{} 1}\frac {1}{n} > \sum_{k \equal{} 1}^{\infty}\sum_{n \equal{} 9\cdot 10^k}^{10^{k \plus{} 1} \minus{} 1}\frac {1}{10^{n \plus{} 1}} \equal{} \sum_{k \equal{} 1}^{\infty}\frac {10^{n}}{10^{n \plus{} 1}} \equal{} \sum_{k \equal{} 1}^{\infty}\frac {1}{10} \equal{} \infty$. Therefore, it diverges.[/hide]

For #5, can't you just say that since Cantor middle-third set has the same cardinality as the real numbers, it cannot form bijection with natural numbers? (its proof is essentially the same as yours)

So I was looking at the site and noted that Wolfram Research is a sponsor.

Yeah, I got an email from someone in charge of the sponsorships group at Wolfram Research. She wanted to know if I was attending, and gave me a bit of info on what they were trying to do.

Congrats Yubo and Oliver for winnng Georgia Tech.

It is late now.

Will discuss about the problems tomorrow.

* irrelevant question about befuddler's initials *

Disclaimer : Problems not in correct order

1.  Prove that $ {2^{2m} \plus{} 1}/5$ is composite

if m is odd and > 3

[hide]
I used the identity of factorising $ a^4 \plus{} 4b^4$ to do this problem

Please post any other methods
[/hide]

I am very slow in typing and latexing

will post the questions one by one

Btw i used Cauchy's lemma for the inequality problem which the corrector probably did not understand or probably there was a mistake.

#5: Prove
\[ \sum_{i \equal{} 0}^{1111} \frac {1}{\sqrt {9i \plus{} 1} \plus{} \sqrt {9i \plus{} 4}} > 10
\]
#1: ABC is a triangle. E is on AB such that AE=(1/2)EB. Given AB=6,AC=4,BC=5, find CE. (Evidently, from what I heard, lots of people have difficulty with computation on this one ;))

#3: Let $ x_i,y_i,z_i$ be sequences with $ x_1,y_1,z_1 > 0$, $ x_i \equal{} y_{i \minus{} 1} \plus{} 1/z_{i \minus{} 1}$, $ y_i \equal{} z_{i \minus{} 1} \plus{} 1/x_{i \minus{} 1}$, $ z_i \equal{} x_{i \minus{} 1} \plus{} 1/y_{i \minus{} 1}$.
(a) Prove all three increase without bound.
(b) Show at least one of $ x_{200},y_{200},z_{200}$ is greater than 20.

[quote="solafidefarms"]#5: Prove
\[ \sum_{i \equal{} 0}^{1111} \frac {1}{\sqrt {9i \plus{} 1} \plus{} \sqrt {9i \plus{} 4}} > 10
\]
[/quote]
[hide]Here's my solution(I'm curious why #5 in this year is so easy!):
Applying the Cauchy-Schwartz's inequality we have:
$ \sum_{i \equal{} 0}^{1111} \frac {1}{\sqrt {9i \plus{} 1} \plus{} \sqrt {9i \plus{} 4}} \geq(11111)^2/sum_{i \equal{} 0}^{1111}{\sqrt {9i \plus{} 1} \plus{} \sqrt {9i \plus{} 4}}$(the equality actually doesn't occur in this inequality!)
Now applying Cauchy-Schwartz's inequality we get:
$ sum_{i \equal{} 0}^{11111}{\sqrt {9i \plus{} 1} \plus{} \sqrt {9i \plus{} 4}}$ = < $ sqrt((18i \plus{} 55555)(11112))$
Now we just need to prove: $ 11111^2$ > $ 10\sqrt ((18.11111.11112/2 \plus{} 55555)(11112))$
This is equivalent to: $ 11111.sqrt{11111} > 10.sqrt((9.11112 \plus{} 5).11112)$. The last step is just a simple inequality when everything is just a number![/hide]

[hide="This is one of many ways to do #5"]
$ \sum_{i = 0}^{1111} \frac {1}{\sqrt {9i + 1} + \sqrt {9i + 4}} = \frac {1}{3} \sum_{i = 0}^{1111} (\sqrt {9i + 4} - \sqrt {9i + 1})$
by multiplying each term by its conjugate.

Thus it is equivalent to prove that

$ \sum_{i = 0}^{1111} (\sqrt {9i + 4} - \sqrt {9i + 1}) > 30$

Let $ S_1 = \sum_{i = 0}^{1111} (\sqrt {9i + 4} - \sqrt {9i + 1})$
Let ${ S_2 = \sum_{i = 0}^{1110} (\sqrt {9i + 10} - \sqrt {9i + 4})}$

Then $ S_1 + S_2 = \sqrt {10003} - \sqrt {1} > 90 = 3 \times 30$

So all we need to do is prove that

$ 2S_1 \geq S_2$

(we will ignore the 1111th index of $ S_1$, since it only adds to the LHS)

We need to show that

$ 2(\sqrt {9k + 4} - \sqrt {9k + 1}) \geq (\sqrt {9k + 10} - \sqrt {9k + 4})$ for $ 1 \leq k \leq 1110$,

or

$ 3\sqrt {9k + 4} \geq 2\sqrt {9k + 1} + \sqrt {9k + 10}$

Since both sides are positive, we can take the square and do some arithmetic to achieve

$ \frac {18k + 11}{2} \geq \sqrt {(9k + 1)(9k + 10)}$

which follows directly from AM-GM

[/hide]

EDIT: Did 3a) as well

[hide="3a)"]

We will prove the statement indirectly; suppose that indeed one of the sequences is bounded.

WLOG let it be $ x_n$.

Then, since $ x_n = y_n + \frac{1}{z_n}$, $ y_n$ must be bounded also.
Similarly, $ z_n$ must also be bounded.

Since all three sequences are bounded, their product, $ x_n y_n z_n$, must also be bounded.

However,

$ x_{n+1} y_{n+1} z_{n+1}$

$ = x_n y_n z_n + \frac{1}{x_n y_n z_n} + ( x+ \frac{1}{x})+( y+ \frac{1}{y} )+( z+ \frac{1}{z} )$

$ > x_n y_n z_n + 6$

for all $ n$, implying that $ x_n y_n z_n$ is not bounded.

Contradiction.

[/hide]

[hide="my method for #5"]Let $ A$ be the l.h.s. Let $ B \equal{} \sum_{n \equal{} 0}^{3333}\frac {1}{\sqrt {3n \plus{} 1} \plus{} \sqrt {3n \plus{} 4}}$.
Lemma 1: $ B > 33$.
Proof: Multiplying the denominator by its conjugate,
$ B \equal{} \sum_{n \equal{} 0}^{3333}\frac {\sqrt {3n \plus{} 4} \minus{} \sqrt {3n \plus{} 1}}{3} \equal{} \frac {\sqrt {10003} \minus{} 1}{3} > \frac {\sqrt {10000} \minus{} 1}{3} \equal{} 33$.
Lemma 2: $ 3A > B$
Proof: Note $ \frac {1}{\sqrt {3n \plus{} 1} \plus{} \sqrt {3n \plus{} 4}} > \frac {1}{\sqrt {3n \plus{} 4} \plus{} \sqrt {3n \plus{} 7}} > \frac {1}{\sqrt {3n \plus{} 7} \plus{} \sqrt {3n \plus{} 10}}$, which implies $ \frac {3}{\sqrt {3n \plus{} 1} \plus{} \sqrt {3n \plus{} 4}} > \frac {1}{\sqrt {3n \plus{} 1} \plus{} \sqrt {3n \plus{} 4}} \plus{} \frac {1}{\sqrt {3n \plus{} 4} \plus{} \sqrt {3n \plus{} 7}} \plus{} \frac {1}{\sqrt {3n \plus{} 7} \plus{} \sqrt {3n \plus{} 10}}$. Therefore,
$ 3A \equal{} \sum_{n \equal{} 0}^{1111}\frac {3}{\sqrt {9n \plus{} 1} \plus{} \sqrt {9n \plus{} 4}} > \sum_{n \equal{} 0}^{1111}\frac {1}{\sqrt {9n \plus{} 1} \plus{} \sqrt {9n \plus{} 4}} \plus{} \frac {1}{\sqrt {9n \plus{} 4} \plus{} \sqrt {9n \plus{} 7}} \plus{} \frac {1}{\sqrt {9n \plus{} 7} \plus{} \sqrt {9n \plus{} 10}} > \sum_{n \equal{} 0}^{9999}\frac {1}{\sqrt {3n \plus{} 1} \plus{} \sqrt {3n \plus{} 4}} \equal{} B$.
Hence, $ A > B/3 > 11 > 10$.

[/hide]

[hide="my method for 3a"]Let $ s_n \equal{} x_n \plus{} y_n \plus{} z_n$. Suppose for some positive integer $ N$, $ x_n < N$ for all $ n$. Since $ x_n \equal{} y_n \plus{} \frac {1}{z_n}$ and $ y_n \equal{} z_n \plus{} \frac {1}{x_n}$, we have $ y_n, z_n < N$ for all positive $ n$. Thus, $ s_n < 3N$ is bounded.
Note $ s_{n \plus{} 1} \equal{} x_n \plus{} y_n \plus{} z_n \plus{} \frac {1}{x_n} \plus{} \frac {1}{y_n} \plus{} \frac {1}{z_n}$, or $ s_{n \plus{} 1} \minus{} s_{n} \equal{} \frac {1}{x_n} \plus{} \frac {1}{y_n} \plus{} \frac {1}{z_n} > \frac {3}{N}$. Thus, $ s_{n}$ is unbounded. Contradiction.
[/hide]

[quote="solafidefarms"]#5: Prove

#1: ABC is a triangle. E is on AB such that AE=(1/2)EB. Given AB=6,AC=4,BC=5, find CE. (Evidently, from what I heard, lots of people have difficulty with computation on this one ;))
.[/quote]
[hide]Really? Anyway, here's my solution:
$ cos B\equal{}(36\plus{}25\minus{}16)/60\equal{}3/4$
Now we have: $ cosB\equal{}EB^2\plus{}25\minus{}EC^2/(40)$ and $ EB\equal{}4$, we get: $ 11\equal{}EC^2$, which means $ EC\equal{}sqrt{11}$[/hide]

[quote="befuddlers"]

Btw i used Cauchy's lemma for the inequality problem which the corrector probably did not understand or probably there was a mistake.[/quote]

@ghjk i think cauchy's lemma is the same as cauchy scwartz

basically my method was the same.

Cauchy's lemma states that $ x^2/a \plus{} y^2/b ..................................$
>or equal to $ (x\plus{}y\plus{}........................)^2/(a\plus{}b\plus{}............................)$
I was pretty shocked i did not place after i got this problem,the power of 2 problem and the cevian problem

Ah well i guess all those who placed did atleast 4 problems

[hide="Problem 3b"]
Consider $ a_n^2 \equal{} (x_n \plus{} y_n \plus{} z_n)^2$
$ a_2^2 \equal{} (x_1 \plus{} 1/x_1 \plus{}.........\plus{} z_1 \plus{} 1/z_1)^2 \geq 36 \equal{} 2.18$
By induction we can prove $ a_{n\plus{}1}^2 \geq a_n^2 \plus{} 18$
$ a_n^2 > 18n$ for $ n>2$
By the way this is not my own solution. I found this in Problem Solving Strategies Chapter 9 Sequences Exercise problem 24. If any of you doesn't have this book and wants a more detailed solution, please tell me. I will post it here.
And i am really surprised that the proof based exams are borrowing questions from such well known books. Even KSU proof exam had a question from the same book. This simply gives an unfair advantage to a person who has already seen this problem. [/hide]

[quote="befuddlers"][hide="Problem 3b"]
And i am really surprised that the proof based exams are borrowing questions from such well known books. Even KSU proof exam had a question from the same book. This simply gives an unfair advantage to a person who has already seen this problem. [/hide][/quote]

[hide]You know what? Making a new problem that is good for such those contests are really hard. Most of the problems now are always derived from some "original problems" with some new insightful ideas added to it. Drawing a problem randomly from a book or any old contests(even from IMO) is the common thing. It tests how hard you work, and if you already saw that problem, will you actually remember how to do it, or you're stuck with it again?? Basically, I don't see any advantage for drawing such problems from the well-known book. Remember, there are thousands of books, and millions of problems are covered in those book! Just ask yourself why you didn't work hard enough to "see" that problem beforehand. Even I, myself, saw the problem #5 on BAMO contest this year before, and remember exactly it's on the APMO contest, but still miss it because I haven't done that problem before(just glance at it).

Anyway, here's my solution for #3:

a)Since the equations are cyclic, W.L.O.G, assume that x_i=max(x_i,y_i,z_i) with every i. Then we have:
x_i>x_(i-1)(since $ x_i>z_i\equal{}x_(i\minus{}1)\plus{}1/y_(i\minus{}1)>x_(i\minus{}1)$). Therefore, we have:
y_(i-1)+1/z_(i-1)>y_(i-2)+1/z_(i-2) (1)
Now since z_(i-1)>x_(i-2) and x_(i-2)>z_(i-2), we have: z_(i-1)>z_(i-2). Therefore, from (1), we easily see that y_(i-1)>y_(i-2)
Therefore, x_i>x_(i-1)>x_(i-2)>....>x_0>0 and similar to other two sequence $ y_i$ and $ z_i$.
Now I just need to prove that x_i is unbounded.
Let x_0,y_0 and z_0 approaches to 0, I have: x_1 approaches to infinity, y_1 approaches to infinity and z_1 approaches to infinity. Since x_i>x_1,y_i>y_1 and z_i>z_1, we easily see that they are all unbounded!

b) We see that $ x_200\plus{}y_200\plus{}z_200\equal{}x_199\plus{}y_199\plus{}z_199\plus{}(1/x_199\plus{}1/y_199\plus{}1/z_199)$
                     $ x_199\plus{}y_199\plus{}z_199\equal{}x_198\plus{}y_198\plus{}z_198\plus{}1/x_198\plus{}1/y_198\plus{}1/z_198)$
Doing exactly the same for x_i+y_i+z_i(i runs down from 199 to 1) and adding all equations together we get:
$ x_200\plus{}y_200\plus{}z_200\equal{}(x_0\plus{}y_0\plus{}z_0)\plus{}(1/x_199\plus{}1/y_199\plus{}1/z_199)\plus{}...\plus{}(1/x_0\plus{}1/y_0\plus{}1/z_0)>
1999^2.9/(x_1\plus{}x_1\plus{}x_2\plus{}...\plus{}x_199\plus{}y_1\plus{}...\plus{}y_199\plus{}z_1\plus{}...\plus{}z_199)\plus{}6$.
Since $ x_1\plus{}x_1\plus{}...\plus{}x_199<199x_200,y_1\plus{}...\plus{}y_199<199.y_200 and z_1\plus{}...\plus{}z_199<199z_200$(because of (a)), we get: $ x_200\plus{}y_200\plus{}z_200>1999.9/(x_200\plus{}y_200\plus{}z_200)\plus{}6$, which means $ (x_200\plus{}y_200\plus{}z_200)^2\minus{}6(x_200\plus{}y_200\plus{}z_200)\minus{}1999.9>0$, which means $ 3x_200>x_200\plus{}y_200\plus{}z_200>137.146$. Dividing both sides by 3, we get that x_200>20.
Hmm, this inequality is quite strict! Good problem though! I'm not really sure about my solution to the unbounded part since it seems to trivialize part b) if my proof is true. :huh: [/hide]

[quote="befuddlers"][quote="befuddlers"]

Btw i used Cauchy's lemma for the inequality problem which the corrector probably did not understand or probably there was a mistake.[/quote]

@ghjk i think cauchy's lemma is the same as cauchy scwartz

basically my method was the same.

Cauchy's lemma states that $ x^2/a \plus{} y^2/b ..................................$
>or equal to $ (x \plus{} y \plus{} ........................)^2/(a \plus{} b \plus{} ............................)$
I was pretty shocked i did not place after i got this problem,the power of 2 problem and the cevian problem

Ah well i guess all those who placed did atleast 4 problems[/quote]

Uh, the highest I know was 4.5 problems (or perhaps more, Oliver?).

[quote="timwu"]

[hide="my method for 3a"]Let $ s_n \equal{} x_n \plus{} y_n \plus{} z_n$. Suppose for some positive integer $ N$, $ x_n < N$ for all $ n$. Since $ x_n \equal{} y_n \plus{} \frac {1}{z_n}$ and $ y_n \equal{} z_n \plus{} \frac {1}{x_n}$, we have $ y_n, z_n < N$ for all positive $ n$. Thus, $ s_n < 3N$ is bounded.
Note $ s_{n \plus{} 1} \equal{} x_n \plus{} y_n \plus{} z_n \plus{} \frac {1}{x_n} \plus{} \frac {1}{y_n} \plus{} \frac {1}{z_n}$, or $ s_{n \plus{} 1} \minus{} s_{n} \equal{} \frac {1}{x_n} \plus{} \frac {1}{y_n} \plus{} \frac {1}{z_n} > \frac {3}{N}$. Thus, $ s_{n}$ is unbounded. Contradiction.
[/hide][/quote]
Wait! How can $ x_n\equal{}y_n\plus{}1/z_n??$ Is it your typo mistake??

[quote="timwu"]Uh, the highest I know was 4.5 problems (or perhaps more, Oliver?).[/quote]

My score is at most 4.6; I did not complete 2b)

@ befuddlers:

I think that someone who received full credit for 4 problems would have placed.
If you are confident that there are no holes in your proof, then perhaps there was a scoring error.

@ghjk:

I'm pretty sure he meant $ x_{n\plus{}1}$.

[quote="azn6021023"]For Problem one on the multiple choice, a direct application of Stewart's Theorem yields the answer.  :jump:  :icecream:  :heli:

Yes I know that Stewart's Theorem results from a double application of the law of Cosines.  But why not do one formula?[/quote]

Yep even #1 on the proof round was a direct application of Stewart's Theorem

But i was so scared i would lose marks so i brute forced the entire darn thing using cosine rule twice.

Can anybody mind checking my solution above for problem 3?? I think it has a serious flaw on part a), but I don't want to confirm myself that I did it wrong, which means I have to go and fix it :oops: . And 1 more questions anybody?? I think we have 5 problems but I just see 4 totally. Hmm... This year must be an easy year!

@ghjk:

2 things...

1. How can you assume that $ x_i \equal{} \text{max}\{x_i,y_i,z_i\}$ for all $ i$?
and
2. I think the question meant unbounded for any $ x_0, y_0, z_0$, so you cannot let them go to 0?

[quote="iniquitus"]@ghjk:

2 things...

1. How can you assume that $ x_i \equal{} \text{max}\{x_i,y_i,z_i\}$ for all $ i$?[quote]

Consider that these equations are cyclic, we just think of it in terms of a system of equations, where we have 3 variables and 3 equations. On the other hand, it's also true if we assume $ x_1 \equal{} max(x_1,y_1,z_1)$, we still can prove that $ x_2 > x_1$, $ y_2 > y_1$ and $ z_2 > z_1$. So continue to do it again and again, we get the result I showed above.
[quote]and

2. I think the question meant unbounded for any $ x_0, y_0, z_0$, so you cannot let them go to 0?[/quote][/quote][/quote]
This one, I actually think my proof has a flaw since if it does, then the lower bound of x_200 in part b) will be infinity. I'll find another proof!(surely the contradiction method works for this one).

By the way, can you post the 2nd problem on this round?

#1: ABC is a triangle. E is on AB such that AE=(1/2)EB. Given AB=6,AC=4,BC=5, find CE.

#2: Prove that $ \frac{2^{2m}\plus{}1}{5}$ is a composite integer if m is an odd integer greater than 3.

#3: Let $ x_i,y_i,z_i$ be sequences with $ x_1,y_1,z_1 > 0$, $ x_i \equal{} y_{i \minus{} 1} \plus{} 1/z_{i \minus{} 1}$, $ y_i \equal{} z_{i \minus{} 1} \plus{} 1/x_{i \minus{} 1}$, $ z_i \equal{} x_{i \minus{} 1} \plus{} 1/y_{i \minus{} 1}$.
(a) Prove all three increase without bound.
(b) Show at least one of $ x_{200},y_{200},z_{200}$ is greater than 20.

#4: Let S={1, 2, 3, 4, ..., 17}. How many subsets of S are there such that each subset has no two consecutive integers?

#5: Prove
\[ \sum_{i \equal{} 0}^{1111} \frac {1}{\sqrt {9i \plus{} 1} \plus{} \sqrt {9i \plus{} 4}} > 10
\]

In some order. Thanks to solafidefarms for providing 1, 3, and 5.

Yeah, #4 was like pretty easy for me. (Except: be sure to count the empty subset!)

[quote="befuddlers"]Nope the max among $ x_n$,$ y_n$ and $ z_n$ change

Take an example and u will get it.[/quote]
Hmm... We don't need to care about the max among those 3 variables! We care about each sequence, which means x_2>x_1,y_2>y_1 and z_2>z_1. In these 3 "new variables", will there exists such a maximum variable? Yes it will. And in what order? Doesn't matter at all! As long as it exists a maximum variables among x_i,y_i and z_i(if you don't believe me, then there is only 1 case that can happen:x_i=y_i=z_i, right??), we can always prove that each sequence is increasing. Simple argument, isn't it??

[quote="ghjk"]
a)

Since the equations are cyclic, W.L.O.G, assume that $ x_i\equal{}\text{max}(x_i,y_i,z_i)$ with every $ i$.

Then we have:

$ x_i>x_{i\minus{}1}$ (since $ x_i >  z_i \equal{} x_{i\minus{}1}\plus{}\frac{1}{y_{i\minus{}1}}>x_{i\minus{}1}$).

Therefore, we have:

$ y_{i\minus{}1}\plus{}\frac{1}{z_{i\minus{}1}}>y_{i\minus{}2}\plus{}\frac{1}{z_{i\minus{}2}}$ (1)

Now since $ z_{i\minus{}1}>x_{i\minus{}2}$ and $ x_{i\minus{}2}>z_{i\minus{}2}$, we have: $ z_{i\minus{}1}>z_{i\minus{}2}$.

Therefore, from (1), we easily see that $ y_{i\minus{}1}>y_{i\minus{}2}$

Therefore, $ x_i>x_{i\minus{}1}>x_{i\minus{}2}>....>x_0>0$ and similar to other two sequence and now I just need to prove that x_i is unbounded.

Let $ x_0,y_0$ and $ z_0$ approaches to $ 0$, I have: $ x_1$ approaches to infinity, $ y_1$ approaches to infinity and $ z_1$ approaches to infinity.

Since $ x_i>x_1$,$ y_i>y_1$ and $ z_i>z_1$, we easily see that they are all unbounded![/quote]

OK so I cleaned up your Latex a bit to make it clearer...

To show your error (That you cannot assume WLOG that $ x_i$ is the maximum):
Your first argument, $ x_i > x_{i\minus{}1}$, is unfortunately false.

Let $ x_1 \equal{} 10, y_1 \equal{} 1, z_1 \equal{} 1$
Then $ x_2 \equal{} 1\plus{}1 \equal{} 2 < x_1$

[quote="iniquitus"]

OK so I cleaned up your Latex a bit to make it clearer...

To show your error (That you cannot assume WLOG that $ x_i$ is the maximum):
Your first argument, $ x_i > x_{i \minus{} 1}$, is unfortunately false.
Let $ x_1 \equal{} 10, y_1 \equal{} 1, z_1 \equal{} 1$
Then $ x_2 \equal{} 1 \plus{} 1 \equal{} 2 < x_1$[/quote]

Interesting.... You're right :oops: . Hmm... But does your example also show the contradiction to what the problem asks(it asks that all 3 sequences are increasing but now we have: x_2<x_1)?? But my solution can be easily fixed by proving that all 3 sequences are increasing in case two of three initial variables are equal! Your example shows the contradiction to what I intend to prove as well. Is the problem's statement wrong?? Or do I keep missing something?

[quote="iniquitus"]Here's an outline of how I proved it:

1. Start indirectly. Assume that one sequence is bounded, and in THIS CASE we can WLOG assume that the series is $ x_n$

2. Proceed to prove that if $ x_n$ is bounded, then $ y_n$ and $ z_n$ are bounded also

3. Since all three sequences are bounded, their product is bounded for all indices also

4. Show that the product can be arbitrarily large, providing a contradiction.

I don't quite understand your question, but the problem asks to prove that all three sequences are UNBOUNDED, not that they are increasing.

For example, the sequence 2, 1, 11, 10, 20, 19,... (minus 1, plus 10)
is not monotonically increasing but it is still unbounded (it can become arbitrarily large)[/quote]
Okay! I got it now now. I misunderstood the problem's statement! Increase without bound only meant unbounded?? I thought it's monotonically increasing too. :lol: . Hmm... I missed this part already if I took this contest! Wait, I can ask for clarification, so it isn't that serious :roll:

Would anyone like to post their solution to the subset one? :D

(Edit)  Here's mine
[hide] Let $ P_n$ be the number of subsets that can be formed when there are $ n$ elements.  Notice that for $ P_{n\plus{}1}$, there are $ n$ elements that can already form as many subsets as $ P_n$.  The extra element we added can form at most $ P_{n\minus{}1}$ subsets, since it can be added to every subset formed by all the elements except the last one.  $ P_{n\plus{}2}\equal{}P_{n\plus{}1}\plus{}P_{n}$, and it is not hard to find that $ P_1\equal{}2, P_2\equal{}3$.  Then recursively adding leads to $ P_{17}\equal{}4181$[/hide]


quadratics
algebra
quadratic formula
special factorizations

Hi just got a worked complex number solution here, but i didn't quite follow (perhaps the worked solution given by the teacher is wrong?): The question is

z^2 + 4(z) + 5 = 0

The worked solution i  have is as follows:

(z+2)^2 + 1 = 0
(z+2)^2 = -1
z+2 = +- (plus and minus) i
therefore z = 2 +- i

I don't the 3rd and 4th lines of the solution. Could someone please help me?

Well, you're taking the square root of both sides...and the square root of -1 is $i$...it must be $\pm$ since you don't know what sign it would be...

P.S.
Shouldn't it be $z=-2 \pm i$?

Two ways of doing it:
[hide="Method 1"]
Completing the square:
$z^2+4z+5=(z+2)^2+1=0$
$(z+2)^2=-1 \implies z=\boxed{-2\pm i}$ -- plus or minus because  you are taking the square root.[/hide]

[hide="Method 2"]
Quadratic formula:
$z=\frac{ -4 \pm \sqrt{4^2-4(1)(5)}}{2(1)}=\boxed{-2 \pm i}$[/hide]

Thats exactly what i thought. It should be -2 +-i ....my teacher must have got it wrong...thanks  guys!


Hello all! Today, i'll introduce a new problem of sequence. I think it's hard for a lot of mathlinkers! :P

Let a, b are 2 real numbers such that $ 0<a,b<4$. We conside the sequence $ a_n$: $ a_0\equal{}a; a_1\equal{}b; a_{n\plus{}2}\equal{}\frac {2(a_{n\plus{}1}\plus{}a_{n})}{\sqrt{a_{n\plus{}1}}\plus{}\sqrt{a_{n\plus{}1}}}$.
Show that this sequence converges, find $ lim a_n$

[quote="thesunflower"]Hello all! Today, i'll introduce a new problem of sequence. I think it's hard for a lot of mathlinkers! :P

Let a, b are 2 real numbers such that $ 0 < a,b < 4$. We conside the sequence $ a_n$: $ a_0 \equal{} a; a_1 \equal{} b; a_{n \plus{} 2} \equal{} \frac {2(a_{n \plus{} 1} \plus{} a_{n})}{\sqrt {a_{n \plus{} 1}} \plus{} \sqrt {a_{n \plus{} 1}}}$.
Show that this sequence converges, find $ lim a_n$[/quote]
To Mods: Please lock this topic, this is a problem from M&Y, issue 3,,2008, and we can't discuss it now


Find the minimum number n such that for any coloring of the integers from 1 to n, inclusive, into two colors, one can fing monochromatic a,b,c and d ( not necessarily distinct) such that a + b + c = d.

[hide="simple proof"]

It's not too hard to see that $ n < 11$.  Proof:

Let 1 be red.  Then $ 1\plus{}1\plus{}1\equal{}3$ must be blue, so $ 3\plus{}3\plus{}3\equal{}9$ is red.  Hence 4 is blue, or otherwise we have $ 1\plus{}4\plus{}4\equal{}9$ using only red numbers.  But $ 1\plus{}1\plus{}9 \equal{} 3\plus{}4\plus{}4 \equal{} 11$, so $ n < 11$.

Red: 1,2,9,10
Blue: 3,4,5,6,7,8

Hence $ n \equal{} 10$ works and must be the maximum.[/hide]

This solution is nice but I thought this problem would somehow use pidgeohole and I would appreciate it if somebody showed me that solution, which could also lead to a generalization.


[hide="Solution."]
=====
In the expansion of $ (2^1\plus{}1)(2^2\plus{}1)\dots(2^{16}\plus{}1)$, we will have terms of the form $ 2^k$, where $ k\geq 0$, $ k\leq 1\plus{}2\plus{}4\plus{}\dots\plus{}16\equal{}31$. Also, any positive integer $ k$ in $ [0,31]$ will be in the expansion by choosing those $ 2^i$ in the binary form of $ k$. So the expression becomes:

$ \frac{(2^0\plus{}2^1\plus{}\dots\plus{}2^{31})\plus{}1}{2^{30}}\equal{}\frac{2^{32}}{2^{30}}\equal{}\boxed{4}$
=====
[/hide]

Meron akong alam mas madali

just multiply the left side by (2-1)  it doesnt matter kasi 1 naman yung multiplier eh..

then using x^2- y^2 = (x+y)(x-y)  you will see finally it will be (2^32 - 1 + 1)/2^30

 :D

regards

[quote="matheas"]
just multiply the left side by (2-1)[...][/quote]

Very nice solution.  :)


The radius of one of two concentric circles is 4 cm long, and the other is 9 cm long. How many centimeters are in the circumference of a circle which is concentric to the given circles and whose radius is equal to the average of the other two radii? Express your answer in terms of $ \pi$.

The radius of the new circle is $ \frac{4\plus{}9}{2}\equal{}6.5$, so the circumference is $ 2\cdot \pi \cdot 6.5\equal{}\boxed{13\pi }$


algebra
polynomial
vector
Rational Root Theorem
algebra unsolved

Prove that for the polynomial
$ P(x)\equal{}x^5\minus{}x^4\minus{}4x^3\plus{}4x^2\plus{}2$
there is no root of $ P(x)$ is the n-th root of rational number

Anybody any ideas?? help me please  :blush:

We assume that the rational number is not $ 0$, since $ 0$ is not a root.  It is well known that if $ p\in\mathbb{Q}$ and $ \sqrt[n]{p}\not\in\mathbb{Q}$, then the numbers $ 1,\sqrt[n]{p},\sqrt[n]{p^2},...,\sqrt[n]{p^{n\minus{}1}}$ form a linearly independent basis for the vector space $ \mathbb{Q}[\sqrt[n]{p}]$.

Thus, considering $ P(\sqrt[n]{p})$ as a number in $ \mathbb{Q}[\sqrt[n]{p}]$, if $ P(\sqrt[n]{p})\equal{}0$ then the coefficients of $ \sqrt[n]{p^k}$ sum up to zero for all natural $ k$.  Now it is obvious that the only possible values of $ n$ is $ 1,2$.  By the rational root theorem, $ n\equal{}1$ gives no solutions.  For $ n\equal{}2$, we have $ P(\sqrt{p})\equal{}p^2\sqrt{p}\minus{}p^2\minus{}4p\sqrt{p}\plus{}4p\plus{}2\equal{}0$.  Thus, we have $ p^2\minus{}4\equal{}0\Rightarrow p\equal{}2$. But, this gives $ \minus{}p^2\plus{}4p\plus{}2\equal{}6$, a contradiction.


Let M, N, P be points on AB, BC, CA of traingle ABC. Let CM', AN' BP' be images of CM, AN and BP in the bisectors of angles C, A, B. Show that if lines CM, AN, BP meeet in a point or are all paralle to one another, thenCM', AN' BP' also meeet in a point or areall paralle to one another

Let M, N, P be points on AB, BC, CA of triangle ABC, Let M1, N1, P1 be images of M, N, P in the midpoints of the corresponding sides of the traingle. Show that if the perpendiuclars to AB,BC, CA  erected at the pints M, N, P meet in a point, then perpendicuars to AB, BC and CA erated at points, N1, P1 alos meet in a point

1. See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=18472[/url] and [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=58872[/url].


induction
function
combinatorics unsolved
combinatorics

Let $n$ be a given integer with $n$ greater than $7$ , and let $\mathcal{P}$ be a convex polygon with $n$ sides. Any set of $n-3$ diagonals of $\mathcal{P}$ that do not intersect in the interior of the polygon determine a triangulation of $\mathcal{P}$  into $n-2$ triangles. A triangle in the triangulation of $\mathcal{P}$ is an interior triangle if all of its sides are diagonals of $\mathcal{P}$. Express, in terms of $n$, the number of triangulations of $\mathcal{P}$ with exactly two interior triangles, in closed form.

I got $n \cdot 2^{n-9}\cdot \binom{n-4}{4}$ for $n \ge 8$.  Is that correct?
I think I have got a proof for that, but it is easy to make mistakes when counting possibilities...

Lemma 1: A polygon with $ n$ vertices has all its sides colored red except one which is blue. Then the number of triangulations of the polygon such that each triangle contains at least one red side is $ 2^{n-3}$.

The proof is by induction on $ n$. $ n=3$ is obvious. Now for $ n>4$ the triangle that contains the blue side must also contain exactly one of the red sides adjacent to it. We get two cases. In each case by removing this triangle we can apply the induction step to get $ 2^{n-4}$ possibilities, for a total of $ 2^{n-3}$.

Lemma 2: A polygon with $ n$ vertices has all its sides colored red except two which are blue. It is known that we have $ k$, respectively $ l$ consecutive red sides ($ k+l=n-2$). Then the number of triangulations such that each triangle contains at least one red side if $ \binom{k+l}{k}$.

The proof may be done by induction on $ n=k+l+2$. If we denote by $ f(k,l)$ the given function then as the triangle containing one of the blue sides must contain one of the red adjacent sides we get $ f(k,l)=f(k,l-1)+f(k,l-1)$. A combinatorial proof is also possible: Assign plus to a triangle that contains one red side from the $ k$ and minus to a triangle that contains one red side from the $ l$. Starting from the blue side, we can visit all triangles (we go from one triangle to the next that shares a diagonal with it and was not already visited). We shall encounter a defining sequence of $ k$ pluses and $ l$ minuses in some order which may be chosen in $ \binom{k+l}k$ ways.

Also note that lemma 1 follows from lemma 2 if we cut the unique triangle containing two consecutive red sides.

These two lemmas provide the key to the solution. Color the sides of the polygon red and the diagonals of the two interior triangles blue. The polygon will be split into the two interior triangles, four regions that have one blue side and one region with two blue sides (all the other sides are red, the last region may be empty). Let the four regions considered have $ a,b,c,d$ red sides and the last region have $ i$ and $ j$ consecutive red sides, $ a+b+c+d+i+j=n$. Then according to the lemmas we have $ 2^{a-2}2^{b-2}2^{c-2}2^{d-2}\binom{i+j}{j}=2^{n-i-j-8}\binom{i+j}{j}$ ways to complete the triangulation.

We also design a plan to count the triangulations. We can firstly find the $ i$ sides then find $ j$ then compute $ a+b$ then $ c+d$ and finally get $ a,b,c,d$. Each triangulation will occur exactly two times (namely, we could swap the two interior triangles and get the same configuration). So let's compute the total number of possibilities following the plan.

Now assume that we've found the two blue sides. We thus know $ k=a+b, l=c+d, i, j$. It is clear that $ a,b$ and $ c,d$ may be chosen in $ k-3, l-3$ ways because $ a,b,c,d\geq 2$. So we get $ (k-3)(l-3)2^{n-i-j-8}\binom{i+j}{j}$ ways.

Now set $ h=i+j$, then $ l=n-h-k$ so $ (k-3)(l-3)=(k-3)(n-h-k-3)=-k^{2}+k(n-h)-3(n-h-3)$. Also $ 4\leq k \leq n-h-4$.

Next, we can place the $ i$ consecutive red sides in $ n$ ways, then we may choose $ 4\leq k\leq n-h-4$. The next $ j$ red sides will form the next group and the remaining will form the last group.
So we shall have a total of $ n\cdot 2^{n-h-8}\binom{h}{i}\sum_{k=4}^{k=n-h-4}(-k^{2}+k(n-h)-3(n-h-3))$ ways. It can be computed as $ n\cdot 2^{n-h-8}\binom{h}{i}\frac{s(s-1)(s+1)}6$ where $ s=n-h-6$ (we may prove that $ \sum_{a+b=s, a,b \in N}ab=\frac{s(s-1)(s+1)}6$). So we get $ n\cdot 2^{n-h-8}\binom{h}{i}\frac{s(s-1)(s+1)}6$. Next as $ i$ may take values from 0 to $ h$ and $ \sum_{i=0}^{h}\binom{h}{i}=2^{h}$, the sum over all $ i$ will be $ \cdot n2^{n-h-8}2^{h}\frac{s(s-1)(s+1)}6=n\cdot 2^{n-8}\binom{s+1}3$. Finally $ s$ may take values from $ 2$ to $ n-6$. And we know $ \sum_{s=0}{m}\binom{s}{3}=\binom{m+1}{4}$. So we get from here that the total sum is $ n\cdot 2^{n-8}(\binom{n-4}{4}-\binom{3}{4})$. Next we need to divide by 2 because every triangulation was computed twice, as we could start with either of the two sets of consecutive red sides.

We get $ n\cdot 2^{n-9}\cdot \binom{n-4}{4}$. The same result as yours :)

My solution avoids calculations almost completely:

First some notation:

Denote the interior triangles by $ A_{1}A_{2}A_{3}$ and $ B_{1}B_{2}B_{3}$  such that the sides $ A_{1}A_{2}$ and $ B_{1}B_{2}$ are facing each other.

There is a tower of (say $ i$) triangles in between $ A_{1}A_{2}$ and $ B_{1}B_{2}$ and there are towers of triangles on each of the remaining sides $ A_{2}A_{3}$, $ A_{3}A_{1}$, $ B_{2}B_{3}$ and $ B_{3}B_{1}$, say consisting of $ j+1,k+1,m+1$ and $ l+1$ triangles (every such tower contains at least one triangle. Otherwise the triangles $ A_{1}A_{2}A_{3}$ and $ B_{1}B_{2}B_{3}$ wold not be interior triangles.)

Now first choose $ A_{3}$. This can be done in $ n$ ways. Furthermore we have to keep in mind that we will count each triangulation twice (interchanging the roles of the two interior triangles.)

Now let us fix values of $ i,j,k,m,l$. The only condition these numbers are required to satisfy is that they should be integers $ \ge 0$ and that the resulting polygon should have $ n$ points. By simple counting this can be seen to be equivalent to the condition
\[ 1+(j+1)+(k+1)+(i+2)+(l+1)+(m+1)+1 = n, \]
i.e.
\[ i+j+k+l+m = n-8. \]
For fixed $ i,j,k,m,l$ satisfying these conditions the number of triangulations can easily be calculated: Each tower of triangles can be constructed from the bottom line by addinga triangle in each step and determining which of the remaining two sides will be the outer side (= side of the given polygon) (the other one has to be the interior side = diagonal of the given polygon), for the topmost triangle both remaining sides have to be outer sides.
(Of course for the triangulation in between the two interior triangles there is no topmost triangle.)
Thus altogether the number of triangulations is
\[ 2^{i}\cdot 2^{j}\cdot 2^{k}\cdot 2^{l}\cdot 2^{m}= 2^{i+j+k+l+m}= 2^{n-8}. \]

However the number of possible choices for $ i,j,k,l,m$ is just the number of partitions of $ n-8$ into a sum of 5 natural numbers, which is just
\[ \binom{n-8+4}{4}= \binom{n-4}{4}. \]
To summarize the number of triangulations of the given polygon with the given properties is
\[ n \binom{n-4}{4}2^{n-8}/ 2 = n \binom{n-4}{4}2^{n-9}. \]


Here is my IMO 2003 trip diary I write for a website that I'm one of the internal members:
http://db.math.ust.hk/articles/imo_charlie/e_imo_charlie.htm
It's more related to the trip than to math, but well you may still find it intersesting  :)

Very interesting. I also was in Japan last summer and your diary remember me that nice period. Charlie, you are not only good at mathematics, but at writing. Congratulation!

Namdung

You all played so much mahjong that word got around pretty quickly it seems :)

That red pillar in the lift lobby sure brings back a lot of memories...

haha... I remembered the time when the polish team wanted to take a photo on the day of the closing ceremony, but they didn't have their national flag, so they borrowed Singapore's flag, folded it a bit and turned it up-side-down...
All of us in the Singapore team was so amused that we took pictures with them too.. all those interesting memories...  :)