AMC USA(J)MO USAMO geometry incenter circumcircle algebra

Have fun!

time limit: 22.5 hours (I'm one eval person)
there wont be a scoreboard for this, just to do problems and have fun and yeah:

1) The sequence an is defined by a_1 = a_2 = 1, a_n+2 = a_n+1 + 2a_n. The sequence b_n is defined by b_1 = 1, b_2 = 7, b_n+2 = 2b_n+1 + 3b_n. Show that the only integer in both sequences is 1.

2) Show that for each n we can find an n-digit number with all its digits odd which is divisible by 5^n.

3) ABC is a triangle with incenter I. Show that the circumcenters of IAB, IBC, ICA lie on a circle whose center is the circumcenter of ABC.

4) Let p(x) be the polynomial (1 - x)^a (1 - x^2)^b (1 - x^3)^c ... (1 - x^32)^k, where a, b, ..., k are integers. When expanded in powers of x, the coefficient of x1 is -2 and the coefficients of x^2, x^3, ... , x^32 are all zero. Find k.

5) Find all sets of 14 or less fourth powers which sum to 1599.

6) X has n members. Given n+1 subsets of X, each with 3 members, show that we can always find two which have just one member in common.

7) A tetrahedron has at most one edge longer than 1. What is the maximum total length of its edges?

8) Do there exist 1985 distinct positive integers such that the sum of their squares is a cube and the sum of their cubes is a square?

9) A license plate has six digits from 0 to 9 and may have leading zeros. If two plates must always differ in at least two places, what is the largest number of plates that is possible?

10) What is the smallest number of colors needed to color 8 boxes of 6 balls (one color for each ball), so that the balls in each box are all different colors and any pair of colors occurs in at most one box.

11) p(x) is a polynomial with integral coefficients. Show that there are no solutions to the equations p(a) = b, p(b) = c, p(c) = a, with a, b, c distinct integers.

12) A tetrahedron has opposite sides equal. Show that all faces are acute-angled.

13) Show that there is a unique polynomial whose coefficients are all single decimal digits which takes the value n at -2 and at -5.

14) Prove that the product of the two real roots of x^4 + x^3 - 1 = 0 is a root of x^6 + x^4 + x^3 - x^2 - 1 = 0.

15) Prove that if n is not a multiple of 3, then the angle  :pi:/n can be trisected with ruler and compasses.

1. [hide]use characteristic equation to find sequences => (2^n-(-1)^n)/3 and 2*3^(n-1)+(-1)^n. If these are to be equal, 2*(-1)^n=3^n-2^(n-1). However, the LHS is obviously even while the RHS is an odd term minus an even term for all positive integral n except for n=1. However, this first term is indeed 1, as said.[/hide]

2. [hide]the only problem I solved last year. ok. base case = n=1. we have 5. wow. it's a multiple of 5. and it has odd digits. wow. ok, inductive step. for some n, if we have a that has all odd digits and is a multiple of 5^n, then I claim one of c*10^n+a will be a multiple of 5^(n+1), where c is an odd digit. well c*10^n+a=5^n(c*2^n+b), where b is a/5^n (and is an integer). So we need to show that for some c which is an odd digit, c*2^n+b is a multiple of 5. As gcd(2^n,5)=1, and we have 5 choices for c, none of which are congruent to each other mod 5, this is obvious. Boom. induction. done. [/hide]

This is a bit hard to read with subscripts, but

11. [hide]Let P(x) = A_n (x^n) + A_(n-1) (x^(n-1)) + ... + A_1 (x) + A_0.

P(a) - P(b)

= A_n (a^n - b^n) + A_(n-1) (a^(n-1) - b^(n-1)) + ... + A_1 (a - b)

= (a - b) J

where J is some integer. This is true because (a - b) is a factor of every term.

But P(a) = b, and P(b) = c, so b - c = (a - b) J. Likewise,

c - a = (b - c) M = (a - b) JM

a - b = (c - a) L = (a - b) JML

Which implies 1 = JML.

J, M, and L are all  :pm: 1 (because they must be integral). Assume without loss of generality that a<b<c. That forces

c - a = (b - c) M   =>   M = -1

c - a = c - b

a = b

which is not allowed, and therefore no solns in distinct integers. [/hide]

You could always use the sub html tag for subscripts...if you want it to look nice.

5)

[hide]

There are none.

Take it mod 16. Fourth powers are 0 or 1 (mod 16)

but 1599 == 15(mod 16).

So you cant have 14 or less 4th powers whose sum is 1599.

QED

[/hide]

14)

[hide]

Let a,b,c, and d be the roots of x^4 + x^3 - 1 = 0

and let s = ab

            t = cd

            p = a + b

            q =  c + d

from viete's formulas:

st = - 1

p+q = - 1

pq + s + t = 0

sq + pt = 0

solving the system (just through substitution) we can get:

s^6 + s^4 + s^3 - s^2 - 1 = 0

which means s = ab is a root of this polynomial, as desired.

QED

[/hide]

6) X has n members. Given n+1 subsets of X, each with 3 members, show that we can always find two which have just one member in common.

[hide]

Suppose the n+1 subsets had no elements in common. This would require 3n+3 elements, which is 2n+3 more than we have, so the subsets must have elements in common.

If the n+1 subsets all had all three elements in common (they were identical) the proof would not be possible, so we must assume that the problem means there are n+1 unique subsets.

Now, assume that the n+1 subsets each have two elements in common, which must be true if there are not two subsets that have just one member in common. Let the first subset be {1,2,a_1}. The second subset must have two members in common with the first. Let the second subset be {1,2,a_2}. The third subset must therefore be {1,2,a_3}, etc, up to the (n+1)th subset which must be {1,2,a_n+1}. This required n+1+2=n+3 elements, which is three more than we have. Therefore, there must be at least two subsets that share only one member in common. Q.E.D.[/hide]

What if some of them have no elements in common?

In the case that some but not all subsets have no elements in common, [hide]the same argument still applies. Let two subsets that have no elements in common be {1,2,a} and {3,4,b}. We need to get to n+1 subsets, and some of the subsets are going to have to have two elements in common with those two subsets. So we might have {1,2,c}, {3,4,d}, {1,2,e}, {3,4,f}, etc. Any combination of subsets like that until we have n+1 unique subsets. However, we've used even more elements, because now we have the (n+1) elements that are distributed one per subset, and 2 more elements for every subset that does not have two elements in common with every other subset... in this case, n+1+2+2=n+5, but in cases with even more subsets that have more elements in common, the total number of elements required would be n+1+2k where k is the number of subsets with "original" elements. k is 1 in the case that all subsets share two elements in common, etc. Anyway, having some subsets with no elements in common and the remainder with 2 elements in common does not damage the proof...[/hide]

8) [hide]Take any n integers a_1, a_2, ... , a_n and suppose that  :Sigma: ak2  = m and  :Sigma: ak3 = n, multiply each a_k by m4*n3, then the sum of the squares becomes m9*n6 which is a cube, and the sum of the cubes becomes m12*n10 a square.[/hide]


inequalities
function
inequalities unsolved

Given positive reals $x,y,z$ with sum of 1 and $k\geq 1$.Prove that $\frac{xy}{\sqrt{k+z^{2}}}+\frac{yz}{\sqrt{k+x^{2}}}+\frac{zx}{\sqrt{k+y^{2}}}\le \frac{1}{\sqrt{9k+1}}$.

It depends on whether the function, whatever that may be for you, is concave or not.  How would [i]you[/i] approach this problem using multivariable Jensen?

Masoud Zargar

I'm not sure if this is right, but...here goes:

By AM-GM,
$\frac{x+y+z}{3}\ge (xyz)^{1/3}$
$\frac{1}{3}\ge (xyz)^{1/3}$
$xyz \le \frac{1}{27}$
Therefore:
$\frac{1}{27x}\ge yz$
$\frac{1}{27y}\ge xz$
$\frac{1}{27z}\ge xy$

We can replace these into the equation because they're greater than the value in the equ. and on the less than side.

$\frac{1}{x\sqrt{k+x^{2}}}+\frac{1}{y\sqrt{k+y^{2}}}+\frac{1}{z\sqrt{k+z^{2}}}\le \frac{27}{\sqrt{9k+1}}$
Since x,y, and z are interchangeable, we just need to prove that they are each less than a third of the RHS.

${\frac{1}{x\sqrt{k+x^{2}}}\le \frac{27}{\sqrt{9k+1}}}$
$27x\sqrt{k+x^{2}}\ge \sqrt{9k+1}$
$529^{2}(k+x^{2}) \ge 9k+1$
$k(529x^{2}-9) \ge 1-529x^{4}$
So, the question:
Is $k \ge \frac{1-529x^{4}}{529x^{2}-9}$?

Not sure exactly how to solve ($RHS>1$) this in a nice way...anybody have any ideas?

[quote="falconwing64"]I'm not sure if this is right, but...here goes:

By AM-GM,
$\frac{x+y+z}{3}\ge (x+y+z)^{1/3}$
$\frac{1}{3}\ge (x+y+z)^{1/3}$
$xyz \le \frac{1}{27}$
Therefore:
$\frac{x}{27}\ge yz$
$\frac{y}{27}\ge xz$
$\frac{z}{27}\ge xy$[/quote]
I am certain that your logic above is incorrect.  Your AM-GM's first lie is incorrect.  The part where you get $\frac{x}{27}\ge yz$, etc. from $xyz \le \frac{1}{27}$ is wrong (you did the algebra incorrectly).

Whoops, sorry about that...must have done my mental comp. too fast... :roll:

However, I think my logic still holds - just got a tad bit messier...see above for edit.

[quote="falconwing64"]
Since x,y, and z are interchangeable, we just need to prove that they are each less than a third of the RHS.

${\frac{1}{x\sqrt{k+x^{2}}}\le \frac{27}{\sqrt{9k+1}}}$[/quote]
${\frac{27}{\sqrt{9k+1}}}\neq\frac{1}{3}RHS$.  :wink:

Who has made sense of this inequality?Help,thanks!

[quote="suweijie"]Given positive reals $ x,y,z$ with sum of 1 and $ k\geq 1$.Prove that $ \frac {xy}{\sqrt {k \plus{} z^{2}}} \plus{} \frac {yz}{\sqrt {k \plus{} x^{2}}} \plus{} \frac {zx}{\sqrt {k \plus{} y^{2}}}\le \frac {1}{\sqrt {9k \plus{} 1}}$.[/quote]
$ \sum_{cyc}\frac {xy}{\sqrt {k \plus{} z^{2}}}\equal{}(xy\plus{}xz\plus{}yz)\sum_{cyc}\frac{xy}{xy\plus{}xz\plus{}yz}\sqrt{\frac{1}{k\plus{}z^2}}\leq$
$ \leq(xy\plus{}xz\plus{}yz)\sqrt{\sum_{cyc}\frac{xy}{(xy\plus{}xz\plus{}yz)(k\plus{}z^2)}}.$
Thus, it's remains to prove that $ \sum_{cyc}\frac{xy}{k\plus{}z^2}\leq\frac{1}{(1\plus{}9k)(xy\plus{}xz\plus{}yz)},$ which true for $ k\geq1,$ but my proof of the last inequality is very ugly.

I think, it's interesting enough: with the same conditions a following inequality :
\[ \frac {xy}{\sqrt {1 \plus{} 7z^{2}}} \plus{} \frac {yz}{\sqrt {1 \plus{} 7x^{2}}} \plus{} \frac {zx}{\sqrt {1 \plus{} 7y^{2}}}\le \frac {1}{4}
\]
is not true ( $ k \equal{} \frac {1}{7}$ ). :wink:


geometry
geometric transformation
rotation

According to our current calendar year, on what day of the week will January 1 fall in the year 3000? And in the year 1,000,000. You must supply both answers!

So which years are leap years and which are not?

[quote="jimli"]this is wierd cause not all 4 years are leap year.  1900 wasn't a leap year.  So are we supposed to think that all are?[/quote]

EDIT: I wasnt thinking

[quote="Hamster1800"]Method for leap years: every 4 years except when the number is divisible by 100 but not by 400.

examples

1992 is a leap year
1900 is not
1991 is not
2000 is a leap year
1600 is a leap year
2100 is not[/quote]
interesting

[quote="gr8sk8r"]why wouldn't be there be a leap every 4 years?[/quote]
the earth's rotation is a little more than a 365 days by about 1/4 of a day but not exactly so some years aren't leap years.

id like to see a solution for this if someone has it!  I hope theres something elegant... chugging through leap years and random moduli seems awfully savage to me  :lol: [/hide]

here goes:

A normal year is 52 weeks and 1 day. A leap year is 52 weeks and 2 days. This is useful to memorize for problems like this.

First let's find out which day 2000 started on, as millenia will be easiest to work with here. 2005 started on saturday, which is 2 days after the thursday which 2004 did. 2003-wed 2002-tues. 2001- mon. 2000-sat. So 2000 started on a saturday.

Every 400 years, there is 1 century divisible by 400, so that means that the number of leap years is:

$\frac{400}4 - 3 = 97$ and the number of normal years is $400-97 = 303$ Thus, the number of offset days is: $97\cdot2 +303 = 194 + 303 = 497$. Putting that in mod 7 gives 0. So every 400 years the offset is neutral.

Thus, the millenium is the same as 2 400s and a 200. Since half of the 200 year spans have a century divisible by 400, there are $\frac 12 + \frac {200}4 - 2 = 48\frac12$ Notice this is also $\frac 12 \cdot 97$. So putting $48\frac12$ in mod 7 gives us $6\frac12$. Thus, each millenium, the offset is alternating 0 and $-1$. Because in the 2000 millenium (even though it actually started on 2001, we're doing this for simplicity) the centuries divisible by 400 landed on 2000, 2400, and 2800, the 2000 offset is neutral

Thus, in the year 3000, January first should fall on a saturday.

Every 2000 years, the offset is 1 (add the two millenia) and since the time until year 1,000,000 is $1,000,000-2000 = 998000$, we have the offset is $-998000/2000 = -499$ Putting this is mod 7, we get 5 so:

In the year 1,000,000, January first will fall on a Friday.

Is this right?

Day of the week is

day = (date + y + y/4 - [y / 100] + [y / 400] + [31 m/ 12]) mod 7
Where y is the year.

0 is Sunday, 1 is Monday, etc.

So 1/1/3000 is Wednesday? 1/1/1000000 is something like Thursday except that I don't care because I won't live that long  :roll:


[hide="My solution"]

We can solve this by using complimentary probability to determine how many four digit numbers are not doublets.

Case 1: No two adjacent digits are the same. There are $ 9 * 9 * 9 * 9 \equal{} 9^4 \equal{} 6561$ numbers that fit this.

Case 2: Three digits in a row are the same at the end of the number. If they're nonzero, we have $ 9 * 8 \equal{} 72$ possible numbers; if they are zero, we have $ 1 * 9 \equal{} 9$ different numbers.

Case 3: Three digits at the beginning of the number are the same. There are $ 9 * 9 \equal{} 81$ ways that this can be achieved.

Case 4: All digits are the same. This gives us $ 9$ numbers.

Our answer is $ 9000 \minus{} (6561 \plus{} 72 \plus{} 9 \plus{} 81) \equal{} 9000 \minus{} 6723 \equal{} 2277$ different doublets.[/hide]

[quote="erw314"]How would you do this?:

A four digit number is called a doublet if any of its digits is the same as only one neighbor. For example, 1221 is a doublet, but 1222 is not. How many doublets are there?

I got 2259. How about you?[/quote]

sorry, mabye I'm not thinking straight, but why isn't 1222 a doublet under that definition?  the second digit, 2, is the same as only one neighbor, the third digit.

But the third digit is the same as both neighbors.

[hide="My Guess"]Maybe 1000 I know its didn't worth hiding.[/hide]

well that depends on your interpretation.

like if $ bbab$ or $ babb$ would coutn as a doublet (the wording implies it should)

and Z-coli; im pretty sure your not overcounting anything. bbcc and ccbb are nonexistent in your cases. in bbac; you say 9 for b and 9 for a; which is true; b cant equal a. you say c has 9 choices. also true; there are no overcounts because saying c has 9 choices is saying it has every # choice except for the number represetned by a; as im pretty sure $ bbab$ is a legit doublet

and Paiev your overcounting $ aaaa$ many times in your solution

Yeah you're right Mike.
I was undercounting, not overcounting, but by the same number...
This is why I should use paper when doing these problems -_-

And $ bbab$ has to be accepted, there's nothing in the problem
stating it's not...only that you can't have $ 3$ neighbors
with the same number.

[quote="Z-coli"]There's no need for complementary counting.  It requires just as many cases, and might even be harder than just doing it straight out[/quote]

I realized that when I went back and looked at this problem this morning. I shouldn't do math when I'm half-asleep  :blush: I redid this problem earlier today and also got 2268.

[quote]and Paiev your overcounting $ aaaa$ many times in your solution[/quote]

Where? I don't see it.

[quote="Paiev"]
[quote]and Paiev your overcounting $ aaaa$ many times in your solution[/quote]

Where? I don't see it.[/quote]

oh wait...nvm...that'll teach me to read before i say anything. and i think 2268 is right now

[quote="Z-coli"]There's no need for complementary counting.  It requires just as many cases, and might even be harder than just doing it straight out:
Case 1: $ abbc$
We can choose $ a$ in $ 9$ ways, $ b$ in $ 9$ ways, and $ c$ in $ 9$ ways.  Hence, there are $ 9^{3}$ possibilities for this case.
Case 2: $ bbac$
There are $ 9$ possibilities for $ b$, and $ 9$ possibilities for each of $ a$ and $ c$.  Once again, we have that there are $ 9^{3}$ possibilites for this case.
Case 3: $ acbb$
Once again, there are $ 9$ possibilities for $ a$ and $ c$, and $ 9$ possibilities for $ b$.  Hence, there are $ 9^{3}$ possibilities for this case.

However, we are undercounting.  We have not counted the possibility $ aabb$.  We have $ 9$ possibilites for $ a$ and $ 9$ possibilities for $ b$.  We thus have $ 81$ possibilities for this case.

We can now sum up our answers, to get a total of $ 3 \bullet 9^{3} \plus{} 81 \equal{} \boxed {2268}$ total ways.[/quote]

Thanks. I redid the problem and found that I miscalculated. That's probably because I created WAY too many subcases for $ abbc$, $ bbac$, $ acbb$, and $ aabb$  :roll:  .

[quote="hasan4444"]So what is the real solution now :huh:  :( ? [/quote]

The answer is $ \boxed{2268}$.


[quote="yhh3002"]Anyone whose doing KSEA know the qualifying score for the money last year?[/quote]

I don't know for sure... because it's not revealed (at least I think so). I looked it up, but still, there are only recepients...

[url]https://www.ksea.org/uploads/BB/20/NMC-Final%20National%20Award%20Recipients%20-Web%20Posting.doc[/url]

I missed one problem last year and was 3rd place in national(maybe there was a tiebreaker for 29/30)... in 7th grade. That entitled me to $\$150$ or so. It was fine, but I will not go there this year because of another important math field day...

I remember it was not as difficult :D but I don't know what I missed.


FTW
blogs
Alcumus
geometry
3D geometry
sphere

I don't know if i'll make another or not but here it is........

1. How many usernames have the word blue in them?

2. How many Mods are there?

3. What user has the most posts? How many? (I'll give you within 50 if the number changes)

4. What is my favorite sport? [hide="hint"]see my profile[/hide]

5. Who has the all-time high rating on FTW?

6. How many Administrators are there?

7. How many emoticons are there?

8. How many blogs are there? (within 10)

9.Who has the highest point total on Alcumus?

PM me your answers.

Note: PLEASE do not PM once for each answer!

may be another in about a week.....

bluecarneal wished me to post my own scavenger hunt, so here it is!

All of the answers can be found on the AoPS website. The questions are listed from easy to difficult.

1) How many of the emoticons move?
2) What is the post amount range for "Hodge Conjecture"?
3) What is THE RA's blog named?
4) Out of the threads in the Games forum, which thread has the most views?
5) How many contributors are there to "Encyclopedia Mathematica"?
6) Which user has the highest "posts per day" amount?
7) What is FantasyLover's nickname?
8) Who owns the blog titled "Frank's blog"?
9) Out of users without an avatar, who has the most posts?
10) What is one of the fifteen numbers?
11) Who is the inventor of the spinning sphere?
12) How many users have, at least at some time in the past, had a spinning sphere avatar?
13) How many users list their location as "illinois"?
14) Who will be the moderator of "The Legend of the Thirteenth Moon -- The Sequel"?
15) r15s11z55y89w21 designed the avatar for one other user of AoPS--which user?

PM your answers to me.


geometry
college
Ross Mathematics Program
FTW

Hurt: -15; Heal: +10

Alameda 100
Albany 100
American Canyon 100
Antioch 100
Belmont 100
Benicia 100
Berkeley 100
Brentwood 100
Burlingame 100
Calistoga 100
Campbell 100
Concord 100
Cupertino 100
Daly City 100
Danville 100
Dublin 100
El Cerrito 100
Emeryville 100
Fairfield 100
Fremont 107
Gilroy 100
Half Moon Bay 100
Hayward 100
Hercules 100
Hillsborough 100
Lafayette 100
Larkspur 100
Livermore 100
Los Altos 100
Los Gatos 100
Martinez 100
Menlo Park 100
Mill Valley 100
Millbrae 100
Milpitas 101
Moraga 100
Morgan Hill 100
Mountain View 100
Napa 96
Newark 100
Novato 100
Oakland 100
Oakley 100
Orinda 100
Pacifica 100
Palo Alto 102
Piedmont 100
Pinole 100
Pittsburg 99
Pleasant Hill 100
Pleasanton 100
Richmond 100
Ross 90
San Bruno 100
San Carlos 100
San Francisco 100
San Jose 100
San Leandro 100
San Mateo 100
San Pablo 100
San Rafael 100
San Ramon 100
Santa Clara 100
Santa Rosa 100
Saratoga 100
Sausalito 100
Sonoma 100
South San Francisco 100
Sunnyvale 100
Tiburon 100
Union City 100
Vallejo 100
Walnut Creek 100

I am very sorry if I left your city or town out; I tried to include as much as I could remember.

And I propose this:
[i][u]Let's get rid of those places without AoPSers first and then beat each other up. [/u][/i]  :rotfl:

[hide="+10 Pleasanton -15 Napa"]
Alameda 100
Albany 100
American Canyon 100
Antioch 100
Belmont 100
Benicia 100
Berkeley 100
Brentwood 100
Burlingame 100
Calistoga 100
Campbell 100
Concord 100
Cupertino 100
Daly City 100
Danville 100
Dublin 100
El Cerrito 100
Emeryville 100
Fairfield 100
Fremont 107
Gilroy 100
Half Moon Bay 100
Hayward 100
Hercules 100
Hillsborough 100
Lafayette 100
Larkspur 100
Livermore 100
Los Altos 100
Los Gatos 100
Martinez 100
Menlo Park 100
Mill Valley 100
Millbrae 100
Milpitas 101
Moraga 100
Morgan Hill 100
Mountain View 100
Napa 81
Newark 100
Novato 100
Oakland 100
Oakley 100
Orinda 100
Pacifica 100
Palo Alto 102
Piedmont 100
Pinole 100
Pittsburg 99
Pleasant Hill 100
Pleasanton 110
Richmond 100
Ross 90
San Bruno 100
San Carlos 100
San Francisco 100
San Jose 100
San Leandro 100
San Mateo 100
San Pablo 100
San Rafael 100
San Ramon 100
Santa Clara 100
Santa Rosa 100
Saratoga 100
Sausalito 100
Sonoma 100
South San Francisco 100
Sunnyvale 100
Tiburon 100
Union City 100
Vallejo 100
Walnut Creek 100
[/hide]

I assume standard rules? (e.g. no double-posting, etc.)

[quote="hunter34"]Where the hell is Dallas?[/quote]

Hmmmmm...I wonder what "[b]Bay Area[/b] Hurt and Heal" might mean...

[quote="Alumen"]Foster City doesn't exist?

San Mateo -15[/quote]

Whoops. Sorry. As I said, I only put those that I could remember, and I having trouble remembering everything. Foster City is added. Alumen, I will assume you heal Foster City.

Also, the rules:
1. No posts by the same person within 20 minutes.
2. You must BOTH hurt AND heal.
3. You may separate your amount into different cities/towns; in other words, you could, for example, do City A -7 City B -3 City C -3 City D -1 City E -1 or City F +4 City G +4 City H +2, etc.
4. From now on, state what you changed.
5. All other "conventional" rules.

[hide="Fremont +8 Berkeley +1 Pleasanton +1 Calistoga -6 Antioch -3 Benicia -1"]Alameda 100
Albany 100
American Canyon 100
Antioch 97
Belmont 100
Benicia 99
Berkeley 101
Brentwood 100
Burlingame 100
Calistoga 94
Campbell 100
Concord 100
Cupertino 100
Daly City 100
Danville 100
Dublin 100
El Cerrito 100
Emeryville 100
Fairfield 100
Fremont 115
Gilroy 100
Half Moon Bay 100
Hayward 100
Hercules 100
Hillsborough 100
Lafayette 100
Larkspur 100
Livermore 100
Los Altos 100
Los Gatos 100
Martinez 100
Menlo Park 100
Mill Valley 100
Millbrae 100
Milpitas 101
Moraga 100
Morgan Hill 100
Mountain View 100
Napa 81
Newark 100
Novato 100
Oakland 100
Oakley 100
Orinda 100
Pacifica 100
Palo Alto 102
Piedmont 100
Pinole 100
Pittsburg 99
Pleasant Hill 100
Pleasanton 111
Richmond 100
Ross 90
San Bruno 100
San Carlos 100
San Francisco 100
San Jose 100
San Leandro 100
San Mateo 100
San Pablo 100
San Rafael 100
San Ramon 100
Santa Clara 100
Santa Rosa 100
Saratoga 100
Sausalito 100
Sonoma 100
South San Francisco 100
Sunnyvale 100
Tiburon 100
Union City 100
Vallejo 100
Walnut Creek 100 [/hide]

I was refering why Dallas isn't on the list.

Maybe because Dallas is about 1500 miles from the Bay Area? (that was an estimate; it's probably way off--sorry)

no seriously, there's a reason why there's a forum called "games and fun factory"
a thread intended to be filled with at least 10 pages of completely useless crap doesnt belong here

It WOULD HAVE gone there...but there's no "Bay Area" division of the Fun Factory...

Anyway,

[hide="Fremont +7 Palo Alto +1 Saratoga +1 Pleasanton +1 Antioch -9 Sonoma -5 Calistoga -1"]Alameda 100
Albany 100
American Canyon 100
Antioch 88
Belmont 100
Benicia 99
Berkeley 101
Brentwood 100
Burlingame 100
Calistoga 93
Campbell 100
Concord 100
Cupertino 100
Daly City 100
Danville 100
Dublin 100
El Cerrito 100
Emeryville 100
Fairfield 100
Fremont 122
Gilroy 100
Half Moon Bay 100
Hayward 100
Hercules 100
Hillsborough 100
Lafayette 100
Larkspur 100
Livermore 100
Los Altos 100
Los Gatos 100
Martinez 100
Menlo Park 100
Mill Valley 100
Millbrae 100
Milpitas 101
Moraga 100
Morgan Hill 100
Mountain View 100
Napa 81
Newark 100
Novato 100
Oakland 100
Oakley 100
Orinda 100
Pacifica 100
Palo Alto 103
Piedmont 100
Pinole 100
Pittsburg 99
Pleasant Hill 100
Pleasanton 112
Richmond 100
Ross 90
San Bruno 100
San Carlos 100
San Francisco 100
San Jose 100
San Leandro 100
San Mateo 100
San Pablo 100
San Rafael 100
San Ramon 100
Santa Clara 100
Santa Rosa 100
Saratoga 101
Sausalito 100
Sonoma 95
South San Francisco 100
Sunnyvale 100
Tiburon 100
Union City 100
Vallejo 100
Walnut Creek 100[/hide]

do you really think putting a bay area tag on this makes it not spam

You take away a total of 15 points from whichever cities you don't like (hurt), and you add a total of 10 points to whichever cities you like. For example,

Say there is City U, City V, City W, City X, City Y, City Z.

City U 100
City V 100
City W 100
City X 100
City Y 100
City Z 100

You could do something like:

City U 110
City V 85
City W 100
City X 100
City Y 100
City Z 100

or something like

City U 100
City V 100
City W 100
City X 95
City Y 110
City Z 90

or even

City U 102
City V 98
City W 101
City X 94
City Y 107
City Z 93

as long as what you heal adds up to 10 and what you hurt adds up to 15. When something reaches 0, it "dies."

[quote="junggi"][quote="r15s11z55y89w21"]What, do I put this directly into the Fun Factory and all sorts of people come and complain why it's only Bay Area? Notice that even within the Bay Area forum, somebody has come and wanted their non-Bay Area city on the list.[/quote]
your argument #1: i cant spam in the fun factory, so i must spam in the bay area
your argument #2: someone trolled this topic, thus it is obviously a topic of worth
and what, are you in such a desperate need of wasting time that you want to hurt antioch for fun

i admit, i do spam often in threads here, but making a thread solely for the purpose of having 10+ pages of useless spam is crossing the line[/quote]

No offense, but not a single thing you mentioned made sense.

"your argument #1: i cant spam in the fun factory, so i must spam in the bay area"

Exactly where did I say that? Notice I have proof for what I said and you don't.

"your argument #2: someone trolled this topic, thus it is obviously a topic of worth
and what, are you in such a desperate need of wasting time that you want to hurt antioch for fun"

Plz define "trolled."
Also, that last statement was simply based on the last post. I, in no way, directed anything specifically to Antioch.
Also, I didn't say that either. And of course, I didn't mean any of those two, if you're going to try to twist logic that way.

May I please ask how 10 pages is different from 1 post if they both only occupy the space of one thread? I do not see how it could be possible that this is worse than an array of many of your posts with unneeded content and inappropriate language.

The purpose of this topic was, obviously, not to spam. I am not a spammer; you can find out from a lot of people that I'm a very serious person. Even if it were spam, it is still much better than the rest of the spam I see littered all over the Bay Area forum. Seriously, just think about it yourself. If you were someone from another state, wouldn't you, after seeing something like this in the Fun Factory, say, "If this was limited to Bay Area, why in the world didn't they just keep it in their forum?" I do not enjoy flaming, hypocrisy, and false accusations. I know I might be wrong; in that case, I leave it up to the moderators to delete this thread. But as of now, I maintain that this thread is fine, and if you will like to spend your time to argue, okay, PM me or have a password game in FTW and we could have a talk.

OH GOD FLAME WAR TIME LOL GOGOGO

[quote="r15s11z55y89w21"]No offense, but not a single thing you mentioned made sense.
[/quote]
lol i find it amusing when some people say "no offense" then follow it with a statement that clearly offends
[quote]
"your argument #1: i cant spam in the fun factory, so i must spam in the bay area"

Exactly where did I say that? Notice I have proof for what I said and you don't.
[/quote]
alright sir
fact 1: i said this thread was a load of spam
fact 2: you did not say anything in response to this
fact 3: you only say that this thread cannot go in fun factory, so it should be here
from 1 and 2, i concluded that you agree that this thread is spam, hence fact 3 becomes more or less "i cant post this spamtopic in the fun factory because it does not belong there; hence, i should post it here"
also i would like to know what "proof" you showed, because i dont see any
[quote]
"your argument #2: someone trolled this topic, thus it is obviously a topic of worth
and what, are you in such a desperate need of wasting time that you want to hurt antioch for fun"

Plz define "trolled."
[/quote]
"to troll" is commonly defined as posting intentionally posting something idiotic/irrelevent/etc in an attempt to draw other posters to respond to it completely seriously, waste their time, etc.
in milder cases, it is similar to [i]joking[/i]
you probably just dont have enough forum experience or not know who hunter34 is, but his posts in this topic are clear attempts at trolling, and he succeeded.
BUT
you cited the example of hunter34, i assume in an attempt to "prove" your position, but of course, your case completely falls apart since you were simply citing [i]jokes[/i] to be something serious.
[quote]
Also, that last statement was simply based on the last post. I, in no way, directed anything specifically to Antioch.
Also, I didn't say that either. And of course, I didn't mean any of those two, if you're going to try to twist logic that way.
[/quote]
of course, i did not mean to imply that you directed anything specifically to that school, and im actually kind of amused that you took it that way
i just meant to show how pointless this thread is
[quote]
May I please ask how 10 pages is different from 1 post if they both only occupy the space of one thread? I do not see how it could be possible that this is worse than an array of many of your posts with unneeded content and inappropriate language.
[/quote]
my view on the community forums is that it should keep a casual tone, make jokes, etc, but while still talking about something useful
[i]this topic is not useful.[/i]
a hurt/heal game is something like
person 1: i heal blah i hurt blah
person 2: i heal blah i hurt blah
person 3: i heal blah i hurt blah
person 4: i heal blah i hurt blah
not only do i not see how this can be fun, it also contributes absolutely nothing to the community
[quote]
I am not a spammer; you can find out from a lot of people that I'm a very serious person. Even if it were spam, it is still much better than the rest of the spam I see littered all over the Bay Area forum.
[/quote]
lol indeed
[quote]Seriously, just think about it yourself. If you were someone from another state, wouldn't you, after seeing something like this in the Fun Factory, say, "If this was limited to Bay Area, why in the world didn't they just keep it in their forum?" [/quote]
what i am saying: we should not have this thread. if you so want to participate in a hurt and heal game, do it in the fun factory, with something inclusive of the overall aops community. unless you really gain some sort of personal satisfaction/pride of seeing your area go up and other areas crash, there is no difference in a bay area themed hurt/heal or any other themed hurt/heal.
edit woops this was supposed to be in a pm

[quote="bowei"]i don't like this thread either. i guess i just dont see the point of games like this. seriously, there are better ways to waste your time than going and playing a stupid forum game like this that could take months to finish with that many cities. there arent that may bay area aopsers anyway who would want to play such a game.

also junggi plz stop trolling/flaming[/quote]

rofl okay okay

[quote="bowei"]i don't like this thread either. i guess i just dont see the point of games like this. seriously, there are better ways to waste your time than going and playing a stupid forum game like this that could take months to finish with that many cities. there arent that may bay area aopsers anyway who would want to play such a game.

also junggi plz stop trolling/flaming[/quote]
bowei, i like your hat.

oh hey i had some new pics of me/other ppl with my hat
taken at scibowl.

[quote="junggi"]no seriously, there's a reason why there's a forum called "games and fun factory"
[/quote]

Correction: The missouri forum. This is low even compared to games and fun factory.

Can we stop spamming this thread? If this is going to continue, can a moderator please lock this please?

You make this kind of useless thread and expect people to take it seriously? Please...
I invited ccy to come get rid of this.
Last on topic comment: BOWEI PLEASE TELL ME WHERE I CAN GET A HAT LIKE YOURS

^Hypocrisy at its finest =(  You guys dislike the thread, deeming it to be "spam," yet you continue to spam and troll it....

WHAT WE ARE LEGITLY CONTRIBUTING TO THIS THREAD BY
ALERTING PEOPLE TO THE AWESOMENESS OF BOWEI'S HAT


probability
topology
geometry
calculus
search
logarithms
set theory
\/closed

Does anyone recommend any texts to introduce me to rigorous probability theory?  I have already had a little bit of exposure to set theory and basic topology of the reals.  I'm looking for a challenging text that is still accessible to a motivated high school student.

It's just that I'm okay with working naïvely as long as I know that I could prove something rigorously if I ever needed to (but it would take more work).  I do not have this kind of soundness of mind with probability, so I would like to know the best text to pick that up with.

For reference, the kind of thing that bugs me is those probability questions that you're supposed to "use geometry" on.  Like real numbers "chosen at random" from an uncountable set.  It's a mask for calculus, it seems, and I'd like to learn probability.

Without more information on exactly what your background is, it's hard to make a definitive recommendation.

Nevertheless, here are two of my favorites that you might examine.  I don't think these will help you solve the geometric probability problems, but they are wonderful introductions to mathematical probability theory.

Random Walks and Electrical Networks, by Doyle and Snell.  This is now available for free download on the Internet, I think it is at arXiv.  A quick search will turn it up.

@Book{lesigne05,
  author =       {Emmanuel Lesigne},
  title =        {Heads or Tails: An Introduction to Limit Theorems in Probability},
  publisher =    {American Mathematical Society},
  year =         2005,
  volume =       28,
  series =       {Student Mathematical Library},
  ISBN =         {0-821-83714-1},
  annote =       {A very nice introduction to the theory of
                  probability.  All of the analysis is rigorous, but
                  done without formal measure theory.  Has proofs of
                  the Weak Law, Large Deviations, Moderate Deviations,
                  Central Limit Theorem, Law of the Iterated
                  Logarithm, ArcSine Law all in the contest of
                  binomial random variables. A book that I wish I
                  could have written.}
}

Here's two others that I have in the database:

@Book{grimmett01-one-thous-exerc-probab,
  author =       {Geoffrey Grimmett and David Stirzaker},
  title =        {One Thousand Exercises in Probability},
  publisher =    {Oxford University Press},
  year =         2001,
  ISBN =       {0-198-57221-2},
  abstract =       {From the advertising blurb: ``This is an updated and
greatly expanded version of an already well-established and popular
exercise manual.  It provides a wide-ranging selection of
illuminating, informative and entertaining problems, together with
their solution.  Topics include modeling and many applications of
probability theory as well as theoretical aspects.  There are
questions at all ability levels, the majority being of elementary or
intermediate standard.}
}

@Book{meester03-natural-introd-probab-theor,
  author =       {Meester},
  title =        {A Natural Introduction to Probability Theory},
  publisher =    {Birkhauser},
  year =         2003,
  ISBN =         {3-7643-2188-1},
  abstract =       {I have looked at this book very briefly, and it looks
interesting.  It seems to approach probability theory in the way that
I do for this course,, that is, intuitively, leading from elementary
results towards measure-theoretic axiomatic probability theory.  A book
worth considering for the course.}}

Steve Dunbar
MAA Director, American Mathematics Competitions


How about some help with this one:

A line segment of length 6 is divided at random into three parts with restrictions that the longest part has a length greater than 2 and less than 4 and that one of the remaining parts is twice the other remaining part.  Find the probability that the three lengths of the three parts can serve as the lengths of the 3 sides of a triangle.

The answer is apparently 3/8, but I'm having trouble seeing why.

Rough solution:

The trouble comes when the longest one is longer than 3.  We know the longest one is at least 12/5 long because otherwise, in order to satisfy the 2:1 ratio of the remaining two segments, it would no longer be the longest segment.  So 12/5 < Longest < 4.  That is an interval 8/5 wide.  The interval that makes a triangle is when 12/5 < Longest < 3, which is 3/5 wide.  (3/5)/(8/5) = 3/8.


conics
ellipse
analytic geometry
geometry
perimeter

Let ABC be a triangle with fixed perimeter. Let the coordinates of A be (0,0) and B be (c,0). Show that the third vertex C lies on a common ellipse.

Hint: [hide]Let the two vertices of the triangle be the foci of the ellipse.[/hide]

[hide="Solution"]As darkdieuguerre said, $ A$ and $ B$ are the foci of the ellipse. Now, we can show that the sum of the distance between $ C$ and $ A$ and the distance between $ C$ and $ B$ is a constant by noting that, given a perimeter $ P$, $ AC\plus{}BC\plus{}AB\equal{}P$, or $ AC\plus{}BC\equal{}P\minus{}c$, which is a constant.[/hide](Possibly wrong)

Well, I guess that's correct. I have my proof but I just want to have other solutions.

+) Let p is perimeter of $ \triangle ABC$, we have $ AB \plus{} BC \plus{} CA \equal{} p \Longrightarrow CA \plus{} CB \equal{} p \minus{} c \equal{} const \Longrightarrow$
$ C$ lie on a elipse.

+) Let $ C(x;y)$, We have $ AC \plus{} CB \plus{} AB \equal{} p$
$ \Longleftrightarrow \sqrt {x^2 \plus{} y^2} \plus{} \sqrt {(x \minus{} c)^2 \plus{} y^2}\plus{}c \equal{} p$

$ \Longleftrightarrow \sqrt {(x \minus{} c)^2 \plus{} y^2} \equal{} p\minus{}c \minus{} \sqrt {x^2 \plus{} y^2}$

$ \Longleftrightarrow x^2 \plus{} y^2 \plus{} c^2 \minus{} 2cx \equal{} (p \minus{} c)^2 \minus{} 2(p \minus{} c)\sqrt {x^2 \plus{} y^2} \plus{} x^2 \plus{} y^2$

$ \Longleftrightarrow 2(p \minus{} c)\sqrt {x^2 \plus{} y^2} \equal{} (p \minus{} c)^2 \minus{} 2(p \minus{} c)\sqrt {x^2 \plus{} y^2}$

............
$ \Longleftrightarrow \frac {(x \minus{} \frac {c}{2})^2}{\frac {(p \minus{} c)^2}{4}} \plus{} \frac {y^2}{\frac {p(p \minus{} 2c)}{4}} \equal{} 1$

[quote="guile"]I've tried laboriously but I can't arrive at

$ \frac {(x \minus{} \frac {c}{2})^2}{\frac {(p \minus{} c)^2}{4}} \plus{} \frac {y^2}{\frac {p(p \minus{} 2c)}{4}} \equal{} 1$

There must be something wrong with my solution. Can you fill up the gap?[/quote]

[size=150][color=darkblue]Yes! It's easy. I will help you:[/color][/size]

[color=darkblue]Let $ C(x;y)$, We have $ AC \plus{} CB \plus{} AB \equal{} p$[/color]

$ \Longleftrightarrow \sqrt {x^2 \plus{} y^2} \plus{} \sqrt {(x \minus{} c)^2 \plus{} y^2} \plus{} c \equal{} p$

$ \Longleftrightarrow \sqrt {(x \minus{} c)^2 \plus{} y^2} \equal{} p \minus{} c \minus{} \sqrt {x^2 \plus{} y^2}$

$ \Longleftrightarrow x^2 \plus{} y^2 \plus{} c^2 \minus{} 2cx \equal{} (p \minus{} c)^2 \minus{} 2(p \minus{} c)\sqrt {x^2 \plus{} y^2} \plus{} x^2 \plus{} y^2$

$ \Longleftrightarrow 2(p \minus{} c)\sqrt {x^2 \plus{} y^2} \equal{} (p \minus{} c)^2 \minus{} c^2 \plus{} 2cx$

$ \Longleftrightarrow 2(p \minus{} c)\sqrt {x^2 \plus{} y^2} \equal{} p(p \minus{} 2c) \plus{} 2cx$

$ \Longleftrightarrow 4(p \minus{} c)^2(x^2 \plus{} y^2) \equal{} p^2(p \minus{} 2c)^2 \plus{} 4c^2x^2 \plus{} 4cxp(p \minus{} 2c)$

$ \Longleftrightarrow 4(p \minus{} c)^2x^2 \plus{} 4(p \minus{} c)^2y^2 \minus{} 4c^2x^2 \minus{} 4cxp(p \minus{} 2c) \equal{} p^2(p \minus{} 2c)^2$

$ \Longleftrightarrow \left[ 4(p \minus{} c)^2x^2 \minus{} 4c^2x^2 \minus{} 4cxp(p \minus{} 2c)\right] \plus{} 4(p \minus{} c)^2y^2 \equal{} p^2(p \minus{} 2c)^2$

$ \Longleftrightarrow \left[ 4x^2p(p \minus{} 2c) \minus{} 4cxp(p \minus{} 2c)\right] \plus{} 4(p \minus{} c)^2y^2 \equal{} p^2(p \minus{} 2c)^2$

$ \Longleftrightarrow 4p(p \minus{} 2c)\left( x^2 \minus{} cx \right) \plus{} 4(p \minus{} c)^2y^2 \equal{} p^2(p \minus{} 2c)^2$

$ \Longleftrightarrow 4p(p \minus{} 2c)\left( x^2 \minus{} cx \plus{} \frac {c^2}{4} \minus{} \frac {c^2}{4}\right) \plus{} 4(p \minus{} c)^2y^2 \equal{} p^2(p \minus{} 2c)^2$

$ \Longleftrightarrow 4p(p \minus{} 2c)\left( x^2 \minus{} cx \plus{} \frac {c^2}{4}\right) \plus{} 4(p \minus{} c)^2y^2 \equal{} p^2(p \minus{} 2c)^2 \plus{} 4p(p \minus{} 2c).\frac {c^2}{4}$

$ \Longleftrightarrow 4p(p \minus{} 2c)\left( x \minus{} \frac {c}{2}\right)^2 \plus{} 4(p \minus{} c)^2y^2 \equal{} p(p \minus{} 2c)(p \minus{} c)^2$

$ \Longleftrightarrow \frac {(x \minus{} \frac {c}{2})^2}{\frac {(p \minus{} c)^2}{4}} \plus{} \frac {y^2}{\frac {p(p \minus{} 2c)}{4}} \equal{} 1$

[color=darkblue]Thanhnam2902[/color]  :P

Question: In line 5, we have
$ \Longleftrightarrow 2(p \minus{} c)\sqrt {x^2 \plus{} y^2} \equal{} p(p \minus{} 2c) \plus{} 2cx$
Then in line 6:

$ \Longleftrightarrow 4(p \minus{} c)^2(x^2 \plus{} y^2) \equal{} p^2(p \minus{} 2c)^2 \plus{} 4c^2x^2 \plus{} 4cxp(p \minus{} 2c)$

Both sides were squared right? where did $ 4cxp(p \minus{} 2c)$ come from?

He square both sides. and $ (p(p\minus{}2c)\plus{}2cx)^2\equal{}p^2(p\minus{}2c)^2\plus{}2*2cx*p(p\minus{}2c)\plus{}4c^2x^2$ .

Something to notice: This is just the definition (or one of them) of an ellipse. As the second poster said, let $ A$ and $ B$ be the foci of the ellipse.

$ AB$ is clearly constant. So we want $ AC\plus{}CB$ to be constant...

[quote="doabli"]He square both sides. and $ (p(p \minus{} 2c) \plus{} 2cx)^2 \equal{} p^2(p \minus{} 2c)^2 \plus{} 2*2cx*p(p \minus{} 2c) \plus{} 4c^2x^2$ .[/quote]

Oh, yeah i got it. my paper was so messy that i didn't see it right away. thanks!

[quote="Yongyi781"]Something to notice: This is just the definition (or one of them) of an ellipse. As the second poster said, let $ A$ and $ B$ be the foci of the ellipse.

$ AB$ is clearly constant. So we want $ AC \plus{} CB$ to be constant...[/quote]

yeah. actually, the locus of points of $ C$ will be an ellipse since $ c$ is constant and therefore the sum of the remaining two sides is constant.


superior algebra
superior algebra solved

What exactly does this ring look like: Z[x]/(x^2+1, 10x-10).

I realize this is really the Gaussian integers, Z[i]/(10i-10), but I really don't know how this looks since the linear term is not monic.

thanks

By the Chinese remainder thm
$\mathbb Z[i]/(10i-10)\simeq \mathbb Z[i]/(1+i)^{3}\times\mathbb Z[i]/(1+2i)\times\mathbb Z[i]/(1-2i)$. The later two rings each have $5$ elements, so it is not too hard to figure out who they are. The other is a ring with $8$ elements and an element $x$ such that $x^{3}=0$ and $3$ is the least such number.

[edit] I don't know how to accurately describe $\mathbb Z[i]/(1+i)^{3}$.
We could say that it is $\mathbb Z[X]/(X^{2}+1,2X-2)$.
We can see that it has $8$ elements i.e. the norm of $(1+i)^{3}$.
It has characteristic $4$ because $4~(1+i)^{4}$ in $\mathbb Z[i]$.
It is not $\mathbb Z/4\times\mathbb Z/2$ because it has a nilpotent of order $3$ nor is it $\mathbb Z/8$ or $(\mathbb Z/2)^{2}$ because it has characteristic $4$. Yet, it is a commutative ring.
We will have to live with it as it is.
Additively, it is $\mathbb Z/4\times\mathbb Z/2$.
One can give a multiplication table, but I won't.


ceiling function
floor function
logarithms
algebra unsolved
algebra

Find all positive integers n such that $ \lceil \frac{3}{\sqrt[n]{3} \minus{} 1} \rceil \equal{} \lfloor \frac{3n}{\ln 3} \rfloor$

For all n $ \frac{3}{\sqrt[n] 3 \minus{}1}<\frac{3n}{ln 3}\minus{}1$

[quote="Arne"]A related question which I found not so easy: prove or disprove that for some $ n$, $ \left\lceil \frac {2}{\sqrt [n]{2} \minus{} 1} \right\rceil \neq \left\lfloor \frac {2n}{\ln 2} \right\rfloor$. Have fun ;)[/quote]
It is easy question, because $ \frac {2n}{\ln 2} \minus{} 1 < \frac {2}{\sqrt [n] 2 \minus{} 1} < \frac {2n}{\ln 2}$. Therefore there are not n, satisfyed your condition.

That's not the right way to answer the question, Rust. Do you see what's wrong?

Let $ x = \frac {\ln a}{n},a>1$.
$ \frac {a}{\sqrt [n] a - 1} = \frac {a}{e^x - 1} = \frac {an}{\ln a}* \frac {1}{Q},Q = 1 + \frac {x}{2} + ... + \frac {x^k}{(k + 1)!} + ...$
Obviosly $ \frac {x^k}{(k + 1)!} < \frac {x^k}{2^k},k > 1$, therefore $ (1 - \frac x2 )Q < 1$. It give
${ 1 > \frac {1}{Q} > (1 - \frac x2})\to \frac {an}{\ln a} > \frac {a}{\sqrt [n] a - 1} > \frac {an}{\ln a} - \frac a2$.

It is not hard to prove, that $ \frac {1}{e^x - 1} - \frac {1}{x}$ increase for $ x > 0$.

Therefore $ c_n = \frac {a}{\sqrt [n] a - 1} - \frac {an}{\ln a}$ decrease, when $ n\to \infty$ or $ c_1 > c_2 > ... > c_{\infty} = - \frac a2$.

From $ \frac {2n}{\ln 2} \minus{} 1 < \frac {2}{\sqrt [n] 2 \minus{} 1} < \frac {2n}{\ln 2}$ it does not follow that $ \left\lceil \frac {2}{\sqrt [n]{2} \minus{} 1} \right\rceil \equal{} \left\lfloor \frac {2n}{\ln 2} \right\rfloor$.

What if, for example, $ \frac {2n}{\ln 2}$ ends in $ 0.999987\cdots$ and $ \frac {2}{\sqrt [n]{2} \minus{} 1}$ ends in $ 0.0000123\cdots$ or something like that?

Yes. You are right. Minimal contrexample is $ n\equal{}777 451 915 729 368$. Founded in (russian forum dxdy).


MATHCOUNTS
probability
arithmetic sequence

1. what is the units digit when 50542(base 6) is written in base 5?

2. nan has art every 6th day of school. assume that school is held monday thru friday for 36 consec weeks. school starts on a mon and nan has art the first day of school. how many times during the school year will nan have art on monday?

3. there are four red balls and two white balls in a jar. one ball is randomly removed and replaced with a ball of the opposite color. the jar is then shaken and one ball is randomly selected. what is the probability that the ball is red? express your answer as a common fraction.

4. what is the greatest positive integer that must divide the sum of the first ten terms of any arithmetic sequence whose terms are positive integers?

5. fifteen balls are numbered 1-15. four balls are then selected at random without replacement. what is the probability that the sum of the numbers on those four balls is odd? express your answer as a common fraction.

6. if a motorist increases her speed by 8 mph she will cover a distance of 120 miles in 30 fewer min. what is her original rate of speed in mph?

7. a pair of dice must be rolled until a 7 occurs. what is the probability that a 7 is obtained on the 5th toss and not before? express your answer as a decimal to the nearest hundredth.

8. one day i found a receipt from my grandfather's store. the bill was faded so that i couldn't read the price of the gadgets at all, and i could read only 3 of the 5 digits in the total. the bill showed the following:
72 gadgets @ $__ . __ __ each = $ __ 74.5 __
how many dollars did one gadget cost? express your answer as a decimal to 2 decimal places.

9. sixteen cups of water are mixed with 8 cups of pure grape juice. two cups of the mixture are removed and two cups of water are added. the grape juice is now what fractional part of the mixture?

10. when the three integers 618, 343, and 277 are divided by a positive integer, d, where d>1, the remainders are the same. what is the smallest possible value of d?

11. how many digits are there in the expanded form of 2 to the 40th power?

12. if 74 hens lay 74 dozen eggs in 74 days and if 37 hens eat 37 kg of wheat in 37 days, how many kilograms of wheat are needed to produce 1 dozen eggs?

13. in a small town of 100 men, 85 are married, 70 have a telephone, 75 own a car and 80 own their own home. on this basis what is the smallest possible number of men who are married, have their own telephone, own their own car, and own their own home?

14. what is the probability that a randomly chosen factor of 15^90 is a multiple of 15^54? express your answer as a common fraction.

thanks so much!

thanks so much!

Here's the first few:

1)  50542 (base 6)'s ones place is equivalent to ones in base ten, but the tens place is equivalent to "sixes" in our base, and so on.

By multiplying each of the digits by the name of their place and adding

( 2*1+4*6+.....)

We get 6686 (base 10).  Now we must convert to base 5.  To do this, we first find the largest power of 5 less than 6686.  This is 5^5=3125.  We then short-divide 6686 by 3125 and get 2 R 436.  So the first digit of the new number is 2.  The next place stands for multiples of 625, but since 436 is less than 625, we fill this place with a 0.  Then we short-divide by 125 and get 3 R 61.  We continue like this until we wind up with 20321 (base 5).  The units digit of this is [b]1[/b].

2)  The x's represent days Nan has art:

Wk|Day
   |m|t|w|h|f
1  x
2     x
3        x
4            x
5               x
6  x

And so you can see that Nan has Monday art during weeks 1,6,11.... Since there are 36 weeks, the series ends at week 36, with [b]8[/b] terms.

3)  2/3 of the time, this results in an even mixture of red and white.  1/3 of the time, it results in 5 red to 1 white.  So we set up like so:

2/3(1/2)+1/3(5/6)=[b]11/18[/b].  (notice that the method takes a sort of weighted average of the 2 probabilities).

4)  The sum of this series, with initial term a and diffrence between consecutive terms b, sums to 10a+ 45b (because the series runs a, a+b,a+2b, etc. for a total of 10 a's and the 9th triangular number of b's)

Anyway, this sum factors to 5(2a+9b).  So the answer is [b]5[/b].

More in a few minutes...

1.[hide]50542 in base 6= 2+4x6+5x36+5x1296=6686. Make that into base 5 and get 203221[/hide]

2. [hide]if she has art every 6th day, then she has art on a monday once every 6 weeks. 36/6=6[/hide]

3. [hide]Two situations. if you pick a red ball (2/3), then you have a 1/2 chance of getting a red ball again. If you pick a white ball(1/3), then you have a 5/6 chance of getting a red ball. 2/3x1/2+1/3x5/6=1/3+5/18=11/18[/hide]

Now for #2 I am getting weeks 1, 7, 13,...31, 6 terms.

i got 6 too but the answer is 7 somehow...

Come on guys -- [quote]school is held monday thru friday[/quote]

Well, here are some others:

5) The sum of the balls is odd iff the number of odd balls is odd.  There are 7 evens and 8 odds.  There can be 3 odds or 1 odd.  The number of choices for the 3-odd case is C(8,3)*C(7,1), and the number of choices for the 1-odd case is C(7,3)*C(8,1).  There are C(15,4) choices altogether. We evaluate:

[C(8,3)C(7,1)+C(7,3)C(8,1)]/C(15,4) and get [b]32/65[/b].

7) The probability that a 7 is rolled is 1/6, and that of it not being rolled is thus 5/6.  We solve:

(5/6)^4*1/6 :approx: [b].08[/b].

8) By short-dividing the total by 72 and using each possible value for the total's hundreds digit, we see that, when placing a 0 in the empty slot at the end of the total, a pattern emerges with the remainder:
   R
1 26
2 54
3 08
4 36
5 64
And here we see that the remainder can be made a multiple of 72-that is, 72- by increasing the number by some number that can be placed in the empty slot.  Dividing the number 574.50 out by hand, we see that the remainder in the next to last step ( for the tenths digit) is 57.  The multiple of 72 closest to ten times this is 576, or 8*72.  We must add 6 to reach this.  So the gadgets total is 574.56, and one gadget costs [b] 7.98[/b].

u guys, for #2, the answer is 6 because it is every 6 school days that you have an art class, so on the first week you have monday, then tuesday the next week, then wednesday, and so on, but when you reach friday, the next week you don't have art, so you have to skip a week, and therefore, you get 6 days of art on monday

12. if 74 hens lay 74 dozen eggs in 74 days and if 37 hens eat 37 kg of wheat in 37 days, how many kilograms of wheat are needed to produce 1 dozen eggs?

well, that means that 1 hen lays 1dozen eggs, in 74 days. and that 1 hen eats 1 kg, in 37 days, so (1 hen eats 2kg of wheat in 74 days)
1 hen lays 1dozen eggs eats 2 kg of wheat, in 74 days
so 2kg of wheat produces a dozen eggs

11. how many digits are there in the expanded form of 2 to the 40th power?

i believe that since this is only to the 40th power, this approximation is close enough: notice that 2^10 is 1024, that is close to one thousand, so we can say that 2^20 is close to one million, and 2^30 is close to one billion, and 2^40 is close to one trillion, and since this is only to the 40th power, there shouldn't be another digit added on: i got [b]13[/b]

or log_2(40)+1=13

do you want me to show you the proof or shall I give you all time to work on it?  btw, you should know that log_2(1) :approx: .301

[quote]
 im confused about number 8 and the explanation.[/quote]

By dividing the number _74.5_ by 72 and putting a 0 in the slot at the end of the number, while using every possible value for the first slot, starting at 1 and working up, we see that the remainder increases by 28 each time.  When the remainder reaches 64, we see that if we increase the number 574.50 by a bit, we will have a remainder of 72, which is the same as a remainder of 0.  We add .06 and get 574.56.  574.56/72 equals the cost of a gadget, [b]$7.98[/b]

13.  Let us try to eliminate as many men as possible from the smallest group, the 70 men with phones.  First, we eliminate using the 85 married men, like this:  We assume that the 30 phoneless men are married.  This leaves 55 married men with phones ("phoned" men? "enphoned" men)?

We assume that the 45 men already eliminated all own a car. This leaves only 30 cars.  So we are down to 30 men.  Finally, we assume that the 70 eliminated men all own their own homes.  This leaves only 10 homes for the men we haven't already gotten rid of, so [b]10[/b] is the answer.


Show that if $ x,y,z$ are positive integers, then $ (xy+1)(yz+1)(zx+1)$ is perfect square if and only if $ xy+1$, $ yz+1$, $ zx+1$ are all perfect squares.

Ok Thanks [b]canada[/b]

Is there any other solution?


limit
integration
trigonometry
symmetry
real analysis
real analysis unsolved

Ascertain $ \lim_{a\to\infty}\frac 1a\cdot\int_0^{\infty}\frac {x^2\plus{}ax\plus{}1}{x^4\plus{}1}\cdot\arctan x\ \mathrm {dx}\ .$

$ I(a)\equal{}\int_0^\infty \frac{x^2\plus{}ax\plus{}1}{x^4\plus{}1} \, \arctan x \, dx$ is after substitution $ x\to \frac{1}{x}$

$ \int_0^\infty \frac{x^2\plus{}ax\plus{}1}{x^4\plus{}1} \, \arctan\left(\frac{1}{x}\right)\, dx$ .

So $ 2\, I(a)\equal{}\int_0^\infty \frac{x^2\plus{}ax\plus{}1}{x^4\plus{}1} \, \underbrace{\left[ \arctan x\plus{}\arctan\left(\frac{1}{x}\right)\right]}_{\equal{}\frac{\pi}{2}} \, dx$

$ \Rightarrow I(a)\equal{}\frac{\pi}{4} \left[\underbrace{\int_0^\infty \frac{x^2\plus{}1}{x^4\plus{}1} dx}_{\frac{\pi}{\sqrt{2}}}\plus{}\underbrace{\int_0^\infty \frac{ax}{x^4\plus{}1} dx}_{a\, \frac{\pi}{4}} \right] \Rightarrow \frac{I(a)}{a}\to \left(\frac{\pi}{4}\right)^2$ .

A somewhat different approach:

$ I(a)=\int_0^{\infty}\frac{x\arctan x}{x^4+1}\,dx+
\frac1a\int_0^{\infty}\frac{(x^2+1)\arctan x}{x^4+1}\,dx$

Since $ \int_0^{\infty}\frac{(x^2+1)\arctan x}{x^4+1}\,dx$ converges (the integrand is $ O(x^{-2})$ as $ x\to\infty$), the second term, which is the only part that depends on $ a,$ tends to zero.

That leaves us computing the first integral.

Let $ u=\arctan x$ or $ x=\tan u$ so that $ du=\sec^2u\,du$ to get

$ I=\int_0^{\frac{\pi}2}\frac{u\tan u\sec^2u\,du}{\tan^4u+1}=
\int_0^{\frac{\pi}2}\frac{u\sin u\cos u}{\sin^4u+\cos^4u}\,du$

Flip the interval around - that is, turn $ u$ into $ \frac{\pi}2-u.$ Then

$ I=\int_0^{\frac{\pi}2}\frac{(\frac{\pi}2-u)\sin u\cos u}{\sin^4u+\cos^4u}\,du$

So $ 2I=\frac{\pi}2\int_0^{\frac{\pi}2}\frac{\sin u\cos u}{\sin^4u+\cos^4u}\,du$

Now: $ \sin u\cos u=\frac12\sin2u$

$ \sin^4u+\cos^4u=(\sin^2u+\cos^2u)^2-2\sin^2u\cos^2u=
1-\frac12\sin^2{2u}$

So if $ \theta=2u,$

$ I=\frac{\pi}8\int_0^{\pi}\frac{\sin\theta}{2-\sin^2\theta}\,d\theta
=\frac{\pi}8\int_0^{\pi}\frac{\sin\theta}{1+\cos^2\theta}\,d\theta$

$ =\left.\frac{\pi}8\cdot\left(-\arctan(\cos\theta)\right)\right|_0^\pi=
\frac{\pi}8\left(\frac{\pi}4+\frac{\pi}4\right)=\frac{\pi^2}{16}.$

[My answer does agree with Dr. Doe's. We both used a symmetry idea.]


Let [tex]\displaystyle\left\{P_k(x)\right\}^{\infty}_{k=1}[/tex] be a sequence of polynomials of degree less than 100 which converges pointwisely on the interval  [tex][0,\,1][/tex]. Show that [tex]\displaystyle\left\{P_k(x)\right\}^{\infty}_{k=1}[/tex]  converges uniformly on [tex][0,\,1][/tex].

[quote="JBL"]We are operating in a finite vector space on a compact space, so that should automatically turn pointwise into uniform convergence, should it not?[/quote]

Not sure about which "fancy" theorem you are refering to. Vaguely remember something similar to what you stated. No analysis reference books here.

What if I change from the closed interval [tex][0,\,1][/tex] to the open interval [tex](0,\,1)[/tex]?

A hint:

[hide]

A polynomial of degree < n is uniquely determined by it's values on n different points.

[/hide]

A more explicit hint:

[hide]

Use Lagrange Interpolation.

[/hide]


linear algebra
matrix
modular arithmetic
algebra
polynomial
linear algebra unsolved

Prove that the equation $ X^n \equal{} I_{3}$ have unique  solution in $ M_{3}(Q)$ if and only if $ n \in {6k \minus{} 1,6k \plus{} 1}$

Here's the easy side:

The equation $ X^n\equal{}I$ always has $ X\equal{}I$ as one solution.

If $ n$ is even, then $ X\equal{}\minus{}I$ is another solution.

If $ 3\mid n,$ then $ X\equal{}\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}$ is another solution.

Hence we can concentrate only on $ n\equiv 1\pmod{6}$ and $ n\equiv 5\pmod{6}.$

The converse:
The  $ X$-minimal polynomial $ P$ is a product of cyclotomic polynomials. $ \phi(3)\equal{}2,\phi(4)\equal{}2,\phi(6)\equal{}2$, the others being $ >3$. Then $ P$ is a product of 2 at most of these polynomials: $ x\pm{1},x^2\plus{}1,x^2\pm{x\plus{}1}$; thus if $ p\equal{}inf\{n: X^n\equal{}I\}$ then necessarily $ p\in\{1,2,3,4,6\}$. If $ n\equal{}6k\pm{1}$ then $ n$ isn't a multiple of $ 2,3,4,6$; it remains the case $ p\equal{}1$ and $ X\equal{}I$ is the unique solution.  QED


superior algebra
superior algebra unsolved

I'm going to take the "degree of an element" to mean the degree of the subextension generated by it (I'm not sure if this is right).

First off if $ x^n \minus{} a$ is irreducible in $ K[x]$ then a root $ u$ has degree $ n$ because $ \{1, u^1,u^2,...u^n \minus{} 1\}$ is a basis. Now suppose $ m|n$, then $ u^m$ generates a subfield with basis $ \{1, u^m, u^{m^2},...u^{m^{n/m}}\}$ which has $ n/m$ elements. If this were not the case, and say $ u^{m^l}$ is in $ K$, then $ x^n\minus{}a$ would not be irreducible.

Carlos Jacob:

Let F be an extension field of the field K.

If x^n-a in K[x] is irreducible and u in F is a root of x^n-a and m divides n, then prove that the degree of u^m over K is n/m.

LkNsngth:

I'm going to take the "degree of an element" to mean the degree of the subextension generated by it (I'm not sure if this is right).

First off if x^n - a is irreducible in K[x] then a root u has degree n because \{1, u^1,u^2,...u^n - 1\} is a basis. Now suppose m|n, then u^m generates a subfield with basis \{1, u^m, u^{m^2},...u^{m^{n/m}}\} which has n/m elements. If this were not the case, and say u^{m^l} is in K, then x^n-a would not be irreducible.

 Carlos Jacob:
 How do you proof that the set \{1, u^m, u^{m^2},...u^{m^{n/m-1}}\} is a linearly independent set?????


I'm very glad to find out the great solution for this problem.Here is my proof for it:Put [tex]S_1=\sum a,S_2=\sum_{cyclic}ab,S_3=\sum_{cyclic}abc[/tex].Supposing [tex]P(x)=(x-a)(x-b)(x-c)(x-d)=x^4-S_1x^3+S_2x^2-S_3x+abcd[/tex].Rolle's theorem says me that:[tex]P'(x)[/tex] has 3 positive roots[tex] u,v,w-->P'(x)=4(x-u)(x-v)(x-w)=4x^3-4(u+v+w)x^2+4(uv+vw+wu)x-4uvw(1)[/tex].
Moreover,[tex]P'(x)=4x^3-3S_1x^2+2S+2x-S_3(2)[/tex].Since (1),(2) we have:[tex]S_1=\frac43(u+v+w),S_2=2(uv+vw+wu),S_3=4uvw[/tex].And the given inequal is equivalent to:Let [tex]u,v,w[/tex] be positive numbers s.t.:[tex]u+v+w+4uvw=2[/tex].Prove that:[tex]uv+vw+wu\leq 1[/tex].Solve :Put [tex]m=2u,n=2v,p=2w[/tex] and we have :[tex]m+n+p+mnp=4(*) [/tex]Since (*) we can prove :[tex]mn+np+pm\leq m+n+p[/tex](it's well-known)[tex]<->uv+vw+wu\leq \frac12(u+v+w)\leq 1[/tex](because [tex]u+v+w\leq 2[/tex])
Similarly,we can solve the following inequal:Let [tex]a,b,c,d[/tex] be positive numbers s.t.:[tex]2(\sum_{cyc}ab)+\sum_{cyc}abc=16[/tex].Prove that:[tex]3(\sum a)\geq 2(\sum_{cyc}ab)[/tex]

[quote="dhthstn"]This problem is very similar to VN96-problem6[/quote]
Yes,you're right!
-----------------------
Are you Son??

[quote="keira_khtn"]Similarly,we can solve the following inequal:Let [tex]a,b,c,d[/tex] be positive numbers s.t.:[tex]2(\sum_{cyc}ab)+\sum_{cyc}abc=16[/tex].Prove that:[tex]3(\sum a)\geq 2(\sum_{cyc}ab)[/tex][/quote]
It's Vietnam 1996

could you show me a proof for:

[quote]$m+n+p+mnp=4$(*) Since (*) we can prove : $mn+np+pm\leq m+n+p$ (it's well-known)[/quote]

for me it is not well-known.  i'd like to see an elementary soln (anyway, lagrange multipliers seem to work here)

[quote="cezar lupu"]Perfect_Radio i have posted the inequality which you consider not well-known.
Harazi,even posted a solution for this one. ;)[/quote]

hey, i can't find it

[quote="keira_khtn"]Supposing that:[tex]mn+np+pm\geq m+n+p[/tex] we have :[tex]9mnp\geq 4(mn+np+pm)(m+n+p)-(m+n+p)^3[/tex](a another form of Schus inequal).And we can show that:[tex]m+n+p\leq 3[/tex].It's impossible!(because [tex]m+n+p+m+n+p\leq 4[/tex])[/quote]

ok. thanks. very nice soln  :D

[quote="ehsan2004"]Let $a,b,c,d$ be positive real numbers such that $3(a+b+c+d)+4(abc+bcd+cda+dab)=8$. Prove the inequality $ab+ac+bc+ad+bd+cd\leq 2$.[/quote]

it is very late in the night, so forgive me if i made stupid mistakes.

my problem is: it seems to be no equality case (so the ineq should be strict). moreover, i failed to find numbers s.t. the expression $\to 2$.

my guess is that $ab+ac+bc+ad+bd+cd\leq \dfrac{16}{9}$ holds in the given conditions. (i used my comp to verify this and it looks true)

anyway, i'd like to see a counter-example if it's false or a proof if it's true  (the proof keira_kn doesn't work in this case, since the "well-known" ineq is sharp)

well, i managed to prove it by lagrange multipliers, i think. although for the case $a=0$ i used the comp (very ugly computations).

what do you think? is it solvable by elementary techniques?

I don't know  :?  [b]perfect_radio[/b] :huh:  :wallbash:  :bye:

[quote="keira_khtn"]Can u post ur solution??[/quote]

Well... I just used the very ugly (rarely effective) method called "Lagrange Multipliers" to find local extremum points. I checked them with the computer. (I also checked the neighbourhood to be sure it is a maximum).

I continued my process for the remaining cases:

(i) $d=0$

(ii) $c=d=0$  (If I remember correctly this case gives you the maximum $\frac{16}{9}$)

and, finally,

(iii) $b=c=d=0$   :P


If you go to Stuyvesant high school and you remember, can you please post here and tell me how many questions are on the Stuy Math Placement Test? Thanks! :lol:

Can a mod please lock this thread? It got double posted somehow... thanks :)

You should talk to someone from stuy personally or try to talk to the headcoach of math team or whatever you are suppose to do.
Just don't fill up the forum with this type of post where the answer can be found with a small amount of research.
Be brave to talk to Mr. C or Mr. Geller. Send them more personal email or stuff.

BTW, Murph and Magictone got 3rd in ARML, and relay became our weakness again. needed to find a way fixing this problem


Two bicyclists are seven-eights of the way through a mile-long tunnel when a train approaches the closer end at 40mph. the riders take off at the same speed in opposite directions, and each escapes the tunnel as the train passes them. how fast did they ride?

this is my answer
[hide]35mph[/hide]
this is the book's "supposedly" correct answer
[hide]30 mph.[/hide]

Ok, I also got 30.

[hide="solution"]
The train reaches the tunnel at the same time the biker headed towards it exits the tunnel, so time it takes for train to enter tunnel = x/s, when the length of the tunnel = 8x. Next, the train exits the tunnel the same time the other biker travels the 7x to the exit on the other side at speed s, so the time = 7x/s, and is also equivalent to time it takes train to reach tunnel (=x/s) + time it takes to travel the 8x (= 8x/40), so our equation =

7x/s = x/s + 8x/40

As you can see x's cancel out, and solving for s = 30 [/hide]

[hide]
The other rider with the shorter distance exited the tunnel already and the train has a mile to exit.  The other rider though has now only $ \frac{7-1}{8}=\frac{6}{8}\ \text{mile}$.
Using the $ t = \frac{d}{v}$ and since the times it takes that second rider and the train to exit are the same,
$ \frac{1 \ \text{mile}}{40 \ \text{mph}}= \frac{\frac{6}{8}\ \text{mile}}{x}$
$ \boxed{x = 30 \ \text{mph}}$
[/hide]

I'm also pretty sure that the book has the right answer...

this is where i questioned the book.
isn't the distance the other rider travels 7/8 of mile not 6/8?
why would you subtract 7-1/8?
that's how i got 35 :huh:

You forgot to count the amount of time it took for the bicyclist closer to the end to bicycle out of the tunnel, which took 1/8 of the time. :lol:

[quote="mihail911"]yah...but regardless of the other biker (since they did ride in opposite directions), the the one with the longer route still has 7 not 6 eights of a mile to travel to exit the tunnel.. :|[/quote]

THE TRAIN REACHES THE TUNNEL ONLY WHEN THE FIRST BIKER REACHES THE END OF THE TUNNEL  .BY THAT TIME, THE OTHER BIKER ALREADY WENT 1/8 OF A MILE, thus leaving him with 6/8 of a mile.

NO IM NOT SCREAMING.

Well what makes the $ \frac{6}{8}$ the distance the second rider has left to exit is that they are travlling at the same speed, travelling the [i]same[/i] distances in the [i]same[/i] time.

[quote]THE TRAIN REACHES THE TUNNEL ONLY WHEN THE FIRST BIKER REACHES THE END OF THE TUNNEL .BY THAT TIME, THE OTHER BIKER ALREADY WENT 1/8 OF A MILE, thus leaving him with 6/8 of a mile.

NO IM NOT SCREAMING.
[/quote]

thanks..NOW IT MAKES SENSE! :lol:


function
MATHCOUNTS
geometry
geometric transformation
reflection

I found a few of the Chapter round problems tough, and would like help on easy ways to do them... for most of them, I solved them when I had extra time during the Target Rounds, but here are the two I didn't like:

27. The function f is defined by f (n) = f (n − 1) + f (n − 2). It is also true that f (1) = 3 and f (3) = 10. What is the value of f (6)?

29. A play has two male roles, two female roles and two roles that can be either gender. Only a man can be assigned to a male role, and only a woman can be assigned to a female role. If five men and six women audition, in how many ways can the six roles be assigned?

Can someone help me here?

[quote="hockeyadi23"]I found a few of the Chapter round problems tough, and would like help on easy ways to do them... for most of them, I solved them when I had extra time during the Target Rounds, but here are the two I didn't like:

27. The function f is defined by f (n) = f (n − 1) + f (n − 2). It is also true that f (1) = 3 and f (3) = 10. What is the value of f (6)? We know that for n=3, 10=f(2)+3; f(2) is thus 7, and so on.

29. A play has two male roles, two female roles and two roles that can be either gender. Only a man can be assigned to a male role, and only a woman can be assigned to a female role. If five men and six women audition, in how many ways can the six roles be assigned? I don't know if I got that one right, but I think we have 5*4*6*5*7*6, perhaps divided by 3!?

Can someone help me here?[/quote]

For # 29

[hide]For # 29:

First to choose the 2 males, you have 5x4 choices.  Then to choose the two female roles you have 6x5 choices. The the two remaining roles can go to any of the 7 remaining people, so 7x6 choices.
So in total, 5x4x6x5x7x6 = 25200.

Hope that helps[/hide]

27.[hide]  If you think of the function as generating a sequence, then the nth term of the sequence is the sum of the previous two terms of the sequence.  We are given the 1st and 3rd terms, so we can subtract to obtain the 2nd term.  Then we can work forward to find the 6th term:

$3,\thinspace ?,\thinspace 10,\thinspace ?,\thinspace ?,\thinspace ?$
$3,\thinspace 7,\thinspace 10,\thinspace ?,\thinspace ?,\thinspace ?$
$3,\thinspace 7,\thinspace 10,\thinspace 17,\thinspace 27,\thinspace 44$[/hide]

Thanks... now I feel really stupid for missing the combinations problem...  :blush:  :blush:

Thnks rcv. I couldn't figure that problem out either.

Theres another solution to problem 27.
We can rewrite f(6)=f(5)+f(4)=f(4)+f(3)+f(3)+f(2)=f(3)+f(2)+f(3)+f(3)+f(2)=3*(f(3))+2*(f(2)).  We know f(3) is equal to 10 so f(3)=f(2)+f(1)=10 So f(2)=10-f(1).   So f(2) = 7.  So summing our equation we get 3(10)+2(7)=44.  Nice mathcounts problem.

BTW, I really need to learn LaTe.

What was #1 on the sprint round?

I didn't get number 27 until I got home, and I just guessed 50 on the problem.
I don't really remember that number 29 though. wAs it really on the competition?

number one was like what was the distance between (listed two cities)

and there was a chart
it gave you the answer on the chart...
it was easy

It was E. The way I would do it would be to first unfold the second fold and mirror the drawing. Then, unfold the first downward fold and mirror the bottom half.

The Chapter was pretty hard this year, with the tough and hard problems. If anyone remembers the coins in the Sprint round, I got that one wrong with a stupid calculation error.  :(

I was one off with that one!!! #26 (Fountain Problem) I was dumb and put 36pi as the answer...But then realized I should change it to 164pi...and time was called.

How are you supposed to go about doing number 24....That one has been driving me crazy!

[quote="frost13"][quote="kchande"]I would think that the answer to 2nd problem would be :

5C2*6C2*7C2

which is 5*4*6*5*7*6/(2*2*2).

I think the guy who gave the answer as 5*4*6*5*7*6 forgot to divide by  2*2*2 for number of permutations.

Can someone else answer to see if I am right.[/quote]

According to the answer key posted at mathcounts.org my answer is right :)   Since the two roles for men are not the same role, picking Joe for 1 and Mike for the other would be different than Mike for the first role and Joe for the second.  So no need to divide by 2.[/quote]

That one was permutations...

[quote="shake9991"]I had trouble with the one where you folded the square paper and cut the thing... what was the answer[/quote]

I did it with visualization first, but since I had extra time, I ripped up some paper and did it that way. :D

it wasn't that hard to picture it, most of the choices were impossible and couldn't have been the answer... and there was only like 2 left... i think it was E. I remember it because someone told me that they actually tore up like 3 pieces of paper trying to do it, but gave up and did it in their mind. it was hard to do it without scissors

What I did for that problem was draw the picture on my scratch paper, then reflect it across the x-axis to simulate "unfolding" the second fold.  Then reflect that drawing across the y-axis to simulate "unfolding" the first fold.

[color=indigo]

I got #29 (I like those kind of problems :D ) but got stuck #27. My mom showed me an easy way to do it afterwards so now I'm bonking my head agianst the wall[/color]

[quote="LuCy4EvA"]i missed number 29 just because i couldn't decide whether it's a combination problem or permutation... but i guess wrong[/quote]Me too. :(

... I feel stupid because I got confused abt how much time was left so I started to rush a bit.. and then I got a bunch wrong b/c I made stupid errors... for the folding, a few of my teammates could only do it by tearing up their scrap paper, i found it easier to eliminate to solve that one.

how can you tear it and solve the problem without a scissor, this girl got it wrong because she didn't know that the directions matter.... I thought about it, and thought it was kinda weird to be in the middle of a mathcount thing and you are sitting there tearing up paper. All you have to do is reflect the places where they cut it.... forgot what it was called, but we learn it in fifth grade

Was number 29 permuntation or combination... i can give reasons why it's both.... hate it when i have to guess on whether a problem is permutation or combination... it's like having a debate with yourself.... and I usually get it wrong... i don't get the difference on some problems

I thought the same thing, but what works works, doesn't it... although tearing was a BIG waste of time... :lol:  :lol:

I missed #29, I had it right at first and then checked it as a combination problem then went argh duh when I saw it was wrong and my original was right.

i can't remember wat i did but im not even sure i got those ?'s rite  :? bit the function one was pretty hard. did i even answer it? ........


limit
trigonometry
function
calculus
calculus computations

$ \lim_{x\to\infty}\cos{\sqrt{x\plus{}1}}\minus{} \cos{\sqrt{x}}$

Please give its solution. Its answer is $ 0$. Thank you so much for giving me your time.

We will use the following known identity: $ \cos a-\cos b=-2\sin{\frac{a+b}{2}}\sin{\frac{a-b}{2}}$ to get  $ 0\leq |\cos{\sqrt{x+1}}-\cos{\sqrt{x}}|\leq 2|\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}|}$. Since $ \sin$ is a continuous function and we know that $ \lim_{x\rightarrow +\infty}{\sqrt{x+1}-\sqrt{x}}=\lim_{x\rightarrow +\infty}{\frac{1}{\sqrt{x+1}+\sqrt{x}}=0}$, we get $ \lim_{x\rightarrow +\infty}\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}=\frac{1}{2}\sin{\lim_{x\rightarrow +\infty}{\frac{1}{\sqrt{x+1}+\sqrt{x}}}=\frac{1}{2}\sin 0=0}}$. Therefore, using squeeze theorem, we are done and the answer is $ 0$.


Ohhh its going to great!!! I can feel the heat already! Anyway, the problem:

Find a closed expression for:

$ \left(1+\frac{1}{a}\right) \left(1+\frac{1}{a^2}\right)\left(1+\frac{1}{a^4}\right) \cdots \left(1+\frac{1}{a^{2^{100}}}\right)$

And while im at it, how does one get those brackets bigger in latex? So it looks better... one of the few thing i dont know about it.... :blush:

Oh god im so sorry, i tpyed it in wrong, seems to be hapening a lot lately.... :blush:

To get big brackets, do the following:
[code]\left( 1 + \frac{1}{a} \right) \left( 1 + \frac{1}{a^2} \right) \left( 1 + \frac{1}{a^3} \right) \dots \left( 1 + \frac{1}{a^{100}} \right)[/code]
to get
\[\left( 1 + \frac{1}{a} \right) \left( 1 + \frac{1}{a^2} \right) \left( 1 + \frac{1}{a^3} \right) \dots \left( 1 + \frac{1}{a^{100}} \right)\]

By the way, are you sure this has a closed expression?

[quote="seamusoboyle"]Ohhh its going to great!!! I can feel the heat already! [/quote]You can´t even imagine how hot it is  :mad:  :(

Absolutely!

And once youve seen the idea, youll never forget it, its just so nice! By the way the word catylest has something to do with it. Try multiplying it by something clever.

BTW: we in ireland never get much heat, we think 18 degrees is hot, so were all looking forward to the sun, (which, after Athens, we all know is a mistake, but anyway!) have to get packing :lol:  :lol:  :lol:

[quote="seamusoboyle"]Sorry for double posting but i just realised i typed the problem in wrong, whoops :blush:

Sorry for that, its edited[/quote]

I was starting to wonder... :P

[hide="Solution"]

\begin{eqnarray*}
x &=& \left(1+\frac 1a\right)\left(1+\frac 1{a^2}\right)\cdots\left(1+\frac 1{a^{2^{100}}}\right)\\
\left(1-\frac 1a\right)x &=& \left(1-\frac 1a\right)\left(1+\frac 1a\right)\left(1+\frac 1{a^2}\right)\cdots\left(1+\frac 1{a^{2^{100}}}\right)\\
&=& \left(1-\frac 1{a^2}\right)\left(1+\frac 1{a^2}\right)\cdots\left(1+\frac 1{a^{2^{100}}}\right)\\
&=&\left(1-\frac 1{a^4}\right)\left(1+\frac 1{a^4}\right)\cdots\left(1+\frac 1{a^{2^{100}}}\right)\\
&&\vdots\\
&=& 1-\frac 1{a^{2^{101}}}\\
x &=& \frac{a^{2^{101}}-1}{a^{2^{101}-1}(a-1)}
\end{eqnarray*}[/hide]

Yeah...
[hide]Multiply the whole thing by $\left(1-\frac{1}{a}\right)$.  Then that and the first term make $\left(1-\frac{1}{a^2}\right)$.  That and the next term make $\left(1-\frac{1}{a^4}\right)$.  So on.  So the closed form for the whole expression is
$\frac{\left(1+\frac{1}{a^{2^{101}}}\right)}{\left(1-\frac{1}{a}\right)}$
Which is the same as what joml88 said.
[/hide]

When evaluating that expression we multiplied then divided by a variable.  The resulting expression thus happened to not be defined for $a=1$ whereas the original obviously is defined at that value.

To account for this, you could divide the $a-1$ out of $a^{2^{101}}-1$...but then you would have a not so nice expression.  It's probably best to just leave the formula in either K81o7 or my form and just define the function when $a=1$ since it is so easy to evaluate.

When $a=1$, the expression equals $2^{101}$.


AMC
AIME
geometry
geometric transformation
reflection

AB and BC are mirrors of equal length. Light strikes BC at C and is reflected to AB. After several reflections it starts to move away from B and emerges again from between the mirrors. How many times is it reflected by AB or BC if ∠b = 1.994 degrees and ∠a = 19.94 degrees?

[img]http://www.kalva.demon.co.uk/aime/soln/diag9414.jpg[/img]

At each reflection the two angles x are equal:

[img]http://www.kalva.demon.co.uk/aime/soln/dia9414a.jpg[/img]

I'm not so sure that this is right, but it's what I got.

[color=white]We can start by creating a whole set of extra mirrors the same length as BC, each of them with the angle of b between them and each having one point at B.   Arranged like this, we can see our set of reflections as a straight line.  Our beam of light will exit the mirrors when the distance from the beam to B exceeds the length of BC. After n+1 reflections, the angle CBD (Where D is the intersection of the beam and the current mirror) is n*b. Seeing CBD as a triangle, the greatest that the angle CBD can be is 180-2a, because if it is greater, the side BD will exceed the length of BC.  Setting b to 1.994 and a to 19.94, we can see that 1.994n=180-1.994*20.  Rearranging, we have 1.994*(n+20)=180.  Because 1.994*90=179.46 and 1.994*91=181.454, the greatest value that n+20 can have is 90.  Therfore, n=70, and the number of reflections is n+1=71 [/color]

Welcome, Combinatoricsman,

To hide your solution so that you don't spoil someone else's enjoyment of solving the problem, you use the Spoiler control above your message field in your Post a reply window. That will reliably hide the solution from casual view. Anyone who desires to learn from your solution can read it by dragging to select (highlight) the text you have hidden with the spoiler codes.


I would like to ask you the following question for the problem below:
Is it a famous theorem or a problem from a competition?

It is given an acute angled triangle ABC and a random point P inside it.
Ga, Gb, Gc are respectively the centers of the gravity of the triangles:
BCP, CAP, ABP. Prove that the lines: AGa, BGb, CGc
intersects at a common point.

The statement is true even if ABC is any triangle and P is outside it.

I'm sorry - the problem is not for this section - it is too easy. I solved it in a minutes. Please don't solve it.

$G$ is the centroid of $\triangle ABC$ and let $M$ be the midpoint of $BC.$ $AM$ cuts $GP$ at $P'.$ By Menelaus' theorem for $\triangle GPM$ cut by $\overline{AP'G_a},$ we have

$\frac{\overline{AG}}{\overline{AM}} \cdot \frac{\overline{MG_a}}{\overline{G_aP}} \cdot \frac{\overline{PP'}}{\overline{P'G}}=1 \Longrightarrow \frac{\overline{P'G}}{\overline{P'P}}=-\frac{2}{3} \cdot \frac{1}{2}=-\frac{1}{3}.$

Analogously, we show that $BG_b,CG_c$ pass through the same point $P'.$


linear algebra
matrix
induction
linear algebra unsolved

Let $M$ be a set of matrix $A_{3,3}$ that has elements from ${{-1, 1}}$. Prove the fallowing:
1) If A is From M, prove that $-6\leq det(A)\leq 6$
2) Prove that det(A) is divisible with 4
3) Show that det(A) is from {-4, 0, 4} (only fom this set)
4) If A is from M and $A*B=I_{3}$ (so B is its inverse) the B is $not$ from M
5) Prove that if A is from M, then all the elements of $A^{2006}$ are not equal to $0$
6) Determine the number of elements of the set $M$

If $\ A=(a_{ij})_{1\leq i,j\leq3}$ then $\ detA=p_1+p_2+p_3-p_4-p_5-p_6$ where $\ |p_i|=1$.
Then $\ |detA|\leq\sum_{i=1}^{6}|p_i|=6$.
By adding the third column to the first and to the second, the determinant will have every element from the first two columns even, so
$\ detA=2\cdot2detA'$ where $\ A'$ is a matrix with integer elements.
So $\ 4|detA$ and thus $\ detA\in\{-4,0,4\}$.
Because $\ detA\cdot detA^{-1}=1$ we can't have $\ A,A^{-1}\in M$.
We prove by induction that all element of $\ A^p$ are odd. For p=1 it's trivial. Assume that it is true for $\ A^p$.
If $\ A^p=(a'_{ij})$ and $\ A^{p+1}=(a''_{ij})$ then $\ a''_{ij}=a'_{i1}a_{1j}+a'_{i2}a_{2j}+a'_{i3}a_{3j}$ is an odd number, being a sum of three odd numbers. The induction is complete, so the property it true for $\ A^{2006}$.
Every element from $\ A$ can be chosen from $\ {-1,1}$ so the number of matrices from $\ M$ is $\ 2^9$.


I always see this: I know how to use Latex but how do people actually write the word Latex like you always see it written?

\LaTeX
$\LaTeX$

And it is case sensitive.


Solve the  equation:
$\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-5}}}}=5.$

x=30.
The equation has one root only.
...

How can  you prove  that the  equation has  only  the root  $x=30?$. Thanks

Consider two cases.
If $x<30$ and $x> 30$.
 :P

And then what? Perhaps something like $\sqrt{x-5}<5$, then $x-\sqrt{x-5}<30-\sqrt{x-5}<?$

Just  take  $t=\sqrt{x-\sqrt{x-5}}$ and consider  two  cases: $t>5$, $t<5$  which  will  give  you impossible  result. So  the only case  is  $t=5$ which  gives  $x=30$.


function
probability
calculus
integration
real analysis
advanced fields
advanced fields unsolved

Consider a partition of space $E^n$ into infinite many $n-1$ dimensional subsets such that
the union of these (infinite many) subsets is the whole $E^n$ (for example, the plane is the union of infinite many lines)
Let $S$ denote the set of these subsets. Let $Q \subset E^n$. Then I want to prove
if for each $s \in S$, $s \cap Q$ has a measure zero in $E^{n-1}$, then $Q$ has a measure zero in $E^n$.

In fact, I would prefer a reference than a proof since I guess this should follow from some well-established Theorem.
So all I need to do is to cite that reference.
Thanks!

What you're really looking for here is essentially Fubini's Theorem.

Try integrating the characteristic function of your set $Q$ as an iterated integral.

There are some issues about the measurability of the the cross-sections of $Q$ in the $n-1$-dimensional measure; as a result, you're going to have to state your hypotheses carefully.

I did learn the Fubini Theorem in my probability theory course. But I don't know
what is the characteristic function of the set and interated integral.
Could you elaborate a little more about "integrating the characteristic function of your set  as an iterated integral" ?

Thanks a lot.

The characteristic function of a set (also called the indicator function) is simply the function that is 1 on the set and zero off of it. The integral of such a function is the measure of the set. If you've seen Fubini's Theorem, you've seen iterated integrals - that's what the theorem is about.

The Fubini theory seems to work with the product space and product measure. I can see how it work if  we make partitions of $E^n$ along the coodinates. But it is not so obvious to me if it is an arbitary partition. To be more specific, In my problem, I separate the whole space into infinite many of rays. each ray passes the origin and has different direction.  The intuition is that we should do a polar integration. This is easy in 2D case. But I just don't know how to do it in an arbitary dimensional space.

Thanks!

I told my professor about Fubini Theory. But he is not convinced about it unless I can give a rigorous proof. Can anyone help me out? Here, I separate the whole space into infinite many of rays. each ray passes the origin and has different direction.

Thanks a lot.


Having

$ \begin{tabular}[t]{|c|c|c|c|c|c|}
\hline
a & 15 & 16&17&18&19 \\
\hline
a & 50 & 40&60&50&20 \\
\hline
\end{tabular}$

looks pretty good, but then adding \\ becomes

$ \begin{tabular}[t]{|c|c|c|c|c|c|}
\hline
a\\(b) & 15 & 16&17&18&19 \\
\hline
a & 50 & 40&60&50&20 \\
\hline
\end{tabular}$

How do I fix this?

What are you trying to do?  It doesn't seem clear to me.

Is this the effect you want?
$ \begin{tabular}[t]{|c|c|c|c|c|c|}\hline a & \multicolumn{5}{c|}{}\\
\hline (b) & 15 & 16&17&18&19 \\
\hline a & 50 & 40&60&50&20 \\
\hline
\end{tabular}$

Something like that but without the horizontal line between $ a$ and $ (b).$

So basically from this

$ \begin{tabular}[t]{|c|c|c|c|c|c|} \hline a \\
(b) & 15 & 16 & 17 & 18 & 19 \\
\hline a & 50 & 40 & 60 & 50 & 20 \\
\hline \end{tabular}$

he wants the 15, 16, 17, 18, 19 centered vertically in their cell boxes (and their cell boxes as tall as the cell box containing a and (b)).

Yeah, I posted that before but it generates a large white space which I don't know how to remove it.

Check the wiki page for $ \text{\LaTeX}$.  It should be able to help to some extent.

You can adjust the lengths in the code if you wish
$ \begin{tabular}[t]{|c|c|c|c|c|c|}\hline
\parbox{0.2cm}{\vspace{0.5ex}a\\b\vspace{0.5ex}}
& 15 & 16 & 17 & 18 & 19 \\
\hline a & 50 & 40 & 60 & 50 & 20 \\
\hline
\end{tabular}$


Lol...Our 2007 Michigan MATHCOUNTS Team prepped for Texas.

I hope you guys do awesome!!!!!!!!!!!!!!!!!!!!!

dude you look like such a girl in that picture  :D

oh well you still look better than krit

matching hats? lol
they still dont beat my hat though (jk) XD

2007 Michigan Math Counts team OWNED!  :starwars:


[quote="Quevvy"][quote="slamchess"]tricks and tips for this problem?[/quote]
Separate it into quadrants and then graph it.[/quote]
well, i was gonna do that when i was taking the test, but i thought it would take too long so i just skipped it(with hope i would finish the rest w/enough time to come back to it)...

It's a shift of the equation centered around the origin, so you could just graph that. Or, if you fnow what the graph will basically look like(a square w/diagonals aligned with axes), and see that it passes through (-1,0) and (1,0), making the diagonal length 2. Note:throughout, I was working w/the above mentioned equation.

well if you replace the equation with $|x|+|y|=1$ it doesn't change anything important, that's just a coordinate shift

[quote="kheg_k"]Is it possible to graph these kinds of things on graphing calculator? (TI-83 Plus for example)[/quote]
Yes, there is an abs( ) function for absolute value.
However, the equation you put in would have to be in terms of y, so you would need 2 equations. (One for positive y, one for negaative y)

[quote="slamchess"]tricks and tips for this problem?[/quote]

Well actually, figure out when x=0 and that is when y is 4 or 2 and those are two verticies which determine a diagonal therefore the answer is just 2.......

Its a sprint, so no claclators.

[hide]
Simplify it:

$\\ |x+2|+|y-3|=1\\ |x+2|=1-|y-3|\\ \pm (x+2)=\pm (1-y-3)\\ \pm (x+2)=\pm (-2-y)\\ \pm (x+2)=\pm (y+2)$

The four possibilities then become:

$\\ x+2=2+y\\ y=x\\\hline\\ x+2=-2-y\\ y+2=-x-2\\ y=-x-4\\\hline\\-x-2=2+y\\ y=-x-4\\\hline\\-x-2=-y-2\\ x+2=y+2\\ x=y$.

or something like that.

But the two equations are the same. You could do something like that?[/hide]


Equilateral triangle ABP is drawn inside square ABCD.  E is the midpoint of CD, and a circle is drawn with diameter EP.  Find the ratio of the area inside the square but outside the circle and triangle to the total area of the square.

let each side of the square be $n$. Then the area of the square is $n^2$.
the area of the equilateral triangle is $\frac{n^{2}\sqrt3}{4}$.
the area of the circle is $\pi(\frac{2n-n\sqrt{3}}{4})^2$

so what we are looking for is $\frac{n^2-(\frac{n^{2}\sqrt3}{4}+\pi(\frac{2n-n\sqrt{3}}{4})^2)}{n^2}$

now you simplify and do some more stuff, i guess
wow im tired so i might be wrong. somebody check my work :)

hmm... i got:

$\frac{n^2-(\frac{n-\frac{1}{2}n\sqrt{3}}{2})^2\pi-\frac{1}{4}n^2\sqrt{3}}{n^2}$

... whos right? (the strange thing in the middle of the numerater is supposed to be pi...)

dude, our formulas are equivalent.
i just made it pretty so no complex fractions
xD

[quote="buzzer11"]hmm... i got:

$\frac{n^2-(\frac{n-\frac{1}{2}n\sqrt{3}}{2})^2\Pi-\frac{1}{4}n^2\sqrt{3}}{n^2}$

... whos right? (the strange thing in the middle of the numerater is supposed to be pi...)[/quote]

That's a capitol pi =p

shh! let him believe what he wants to believe

haha, keeyounglee, ur right.  :oops:

LordoftheMorons: how do u type in a lower case pi in latex?

[quote="pianoforte"]$\pi$ not $\Pi$

\pi not \Pi[/quote]

thanks, pianoforte. $\pi$ (just trying it out)
ill fix it in my earlier post.


calculus
derivative
logarithms
calculus computations

$\frac{d}{dx}\ln\ln[t(x)] = \frac{1}{\ln[t(x)]}\frac{d}{dx}\ln[t(x)] = \frac{1}{t(x) \cdot \ln[t(x)]}\frac{d}{dx}t(x)$

and so

$\frac{d}{dx}t(x) = t(x) \cdot \ln[t(x)] \cdot \frac{d}{dx}\ln\ln[t(x)]$.

In your case, $t(x) = u^{v^{w}}$, and so

$\ln\ln[t(x)] = \ln\ln(u^{v^{w}}) = \ln(v^{w}\ln u) = \ln(v^{w})+\ln u = w \ln v+\ln u$

and

$\frac{d}{dx}\ln\ln[t(x)] = \frac{d}{dx}[w \ln v+\ln u] = w' \ln v+\frac{w \cdot v'}{v}+\frac{u'}{u}$.

Finally,

$\frac{d}{dx}\left(u^{v^{w}}\right) = u^{v^{w}}\cdot v^{w}\ln u \cdot \left(w' \ln v+\frac{w \cdot v'}{v}+\frac{u'}{u}\right)$.


[quote="silviu"]Fie ecuatia:$x^{2n+1}+k^3x^3+kx-2=0,$ si $k\in N^*$,K fixat.
a)Sa se arate ca ecuatia are o singura radacina reala $x_{n}$,iar sirul $(x_{n})_{n>0}$ este convergent.
b)Determinati limita sirului $(x_{n})$[/quote]

Write as $ f(Y) = (Y/k)^{2n+1} + Y^3 + Y = 2$.
$f$ is increasing and continuous, hence (a).
$0 < Y_n < 1$, so the $(Y/k)^{2n+1} \to 0$.
If $Y_n = 1 - b_n$ we get $4k^{2n+1} b_n \to 1$

I'm sorry fleeting_guest, I don't quite get your solution.

Oricum, solutia mea la pb.
Fie $f_{n}(x)=x^{2n+1}+k^{3}x^{3}+kx-2$
$f_{n}$-continua si f(0)<0; f(1)>0, deci unicul zerou al functiei de afla in intervalul (0,1).
Sa dem ca sirul $x_{n}$ este crescator.
Avem $f_{n+1}(x_{n})=x_{n}^{2n+1}(x_{n}^2-1)\leq0$, deci $x_{n+1}\in\(x_{n},1)$.
Fie l limita sirului xn. l satisface ecuatia $k^{3}l^{3}+kl-2=0$, care se rezolva usor.

Imi cer scuze pentru aspectul post-ul meu de mai jos, nu-mi dau seama ce s-a intamplat cu secventele de cod Latex.

[quote="vlad"]I'm sorry fleeting_guest, I don't quite get your solution.

Oricum, solutia mea la pb.
Fie $f_{n}(x)$=$x^{2n+1}+k^{3}x^{3}+kx-2$
$f_{n}$-continua si f(0)<0; f(1)>0, deci unicul zerou al functiei de afla in intervalul (0,1).
Sa dem ca sirul $x_{n}$ este crescator.
Avem $f_{n+1}(x_{n})=x_{n}^{2n+1}(x_{n}^2-1)\leq0,%Error. "deci" is a bad command.
x_{n+1}\in\(x_{n},1)$.
Fie l limita sirului xn. l satisface ecuatia $k^{3}l^{3}+kl-2=0$, care se rezolva usor.[/quote]

Ti-am modificat eu postul, pentru că era ilizibil. Trebuie să ai mai mare grijă când şi unde pui $ ... $ (că acolo asta era problema).

[quote="vlad"]I'm sorry fleeting_guest, I don't quite get your solution.[/quote]

My reasoning shows that if $Y_n = kx_n$, then

 $Y_n \to 1$  (this solves the problem, i.e. $x_n \to 1/k$)

and more precisely,

  $Y_n = 1 - \frac{(1+o(1))}{4k^{2n+1}} $.

It is easiest to write $Y_n = 1 - b_n$ and compute with the $b_n$.


Prove:
If $g \in L^\infty$, the operator $T$ defined by $Tf=fg$ is bounded on $L^{p}$ for $1\leq p\leq \infty$.
The operator norm is at most $\parallel g\parallel _\infty$, with equality if measure space is semifinite.

(first part is trivial, but how do I show the equality if space is semifinite? How to construct
a $f$ in $L^{p}$ such that for any $g \in L^\infty$, $\parallel fg\parallel _{p}=\parallel f\parallel _{p}\parallel g\parallel _{\infty}$?)

If semifinite sounds intimidating, can someone give a hint on [0,1] case?

Remember that you're given $g$ first; the $f$ you will construct will depend on $g.$

For any $\epsilon>0,$ there is a set $E$ such that $m(E)>0$ and $|g(x)|>\|g\|_{\infty}-\epsilon$ for $x\in E.$

Let $F$ be a measurable set such that $F\subseteq E$ and $0<m(F)<\infty.$ Let $f=c\chi_{F}.$

I'll let you finish this by explicitly computing both $\|f\|_{p}$ and $\|fg\|_{p}.$


find all integer values of $ a,b,c$ satisfying:

\[ 2a\plus{}3b\plus{}5y\equal{}11\]

\[ 2a \plus{} 3b \plus{} 5c \equal{} 11
\]
We have $ a \equal{} \frac{11\minus{}3b\minus{}5c}{2}$. Thus either $ b$ is odd and $ c$ is even, or the other way around.
Giving solutions:
$ (a,b,c) \equal{} (4\minus{}3k\minus{}5j,2k\plus{}1,2j)$
$ (a,b,c) \equal{} (3\minus{}3k\minus{}5j,2k,2j\plus{}1)$


function
algebra
domain
calculus
calculus computations

Consider the function $y = f(x)=-x^3+9x+8$ on the interval $[0, 4]$.

a)Find the min and max value of f(x) on the interval
b)at x = 0, f(x) is inc, dec, has a pt of inflection, or has a relative max value
c)if x = 0, the fcn is conc up or down, at pt of intersection, or there are no points of inflection...

Here's what I have so far:

$f' = -3x^2 + 9$

set to zero $-3x^2+9=0$

$x = -\sqrt{3}, x = \sqrt{3}$

           x    |   f(x)
          -------------
           0    |   8
           4    |   -20
$-\sqrt{3}$| -2.39
 $\sqrt{3}$|18.39

so for a the min of f(x) is -20 and the max is 18.39

for b)

  -                                  +       +      -
-------|------|------|------
       -1.7    0       1.7

so at x=0 f(x) is increasing???

for c) little confused here please help

Note that $-\sqrt{3}$ is none of your concern, as it is not in the domain of the function.

Yes, the function is increasing in a neighborhood of zero, and what happens at $\sqrt{3}$ is the global ("absolute") maximum.

Determining concavity and points of inflecton requires the use of he second derivative. Have you learned that?


Given a sequence of positive real numbers $x_1,x_2,...,x_n$ with $x_1+x_2+...+x_n = 1$, prove that
$(1+\frac{1}{x_1})(1+\frac{1}{x_2})...(1+\frac{1}{x_n})\geq (1+n)^n$,
the equality sign holds if and only if $x_1=x_2=...=x_n$.

Take $ln$ both sides.
Put $f(x)=ln(1+\frac1x)$
      $f''(x)= \frac{2x+1}{x^2(x+1)^2}$
Then Jensen :D

Thanks a lot!  :)
Both solutions are neat. :idea:

or you can use the inequality

$1+\frac{1}{x}\geq \frac{(n+1)^2}{n^2x+1}$ which is equivalent to $(nx-1)^2\geq 0$

Hence

$\prod (1+\frac{1}{x})\geq \prod \frac{(n+1)^2}{n^2x+1}$

by AM-GM ,

$\prod (n^2x+1)\leq \left(\frac{n^2\sum x +n}{n}\right)^n=(n+1)^n$

Hence $\prod \frac{(n+1)^2}{n^2x+1}\geq \frac{(n+1)^{2n}}{(n+1)^n}=(n+1)^n$


limit
integration
trigonometry
inequalities
calculus
function
calculus computations

$ \lim_{x\to 0^\plus{}} {x\int^1_x \frac{\cos t}{t^2} }dt$

$ \frac{1}{x}\plus{}\frac{1}{2}x\minus{}\frac{3}{2}\leq \int_x^1 \frac{\cos t}{t^2}\ dt\leq 1\minus{}\frac{1}{x}\ (x>0)$,

the limit is 1.

[quote="kunny"]$ \frac {1}{x} \plus{} \frac {1}{2}x \minus{} \frac {3}{2}\leq \int_x^1 \frac {\cos t}{t^2}\ dt\leq 1 \minus{} \frac {1}{x}\ (x > 0)$,

the limit is 1.[/quote]

how the limit becomes 1?

You can use the inequality $ 1 \minus{} \frac {t^2}{2}\leq \cos t\leq 1$.

Just idea: we could integral by parts and triangle inequality.

[quote="kunny"]$ \frac {1}{x} \plus{} \frac {1}{2}x \minus{} \frac {3}{2}\leq \int_x^1 \frac {\cos t}{t^2}\ dt\leq 1 \minus{} \frac {1}{x}\ (x > 0)$,

the limit is 1.[/quote]
$ x(\frac {1}{x} \plus{} \frac {1}{2}x \minus{} \frac {3}{2})\leq x\int_x^1 \frac {\cos t}{t^2}\ dt\leq x(1 \minus{} \frac {1}{x})\ (x > 0)$,

$ \implies 1\plus{}\frac {1}{2}x^2 \minus{} \frac {3}{2}x\leq x\int_x^1 \frac {\cos t}{t^2}\ dt\leq (x\minus{}1)$

$ \implies 1\leq \lim_{x\to 0^\plus{}}\leq \minus{}1$

How becomes 1? :oops:

Sorry :oops:

By Integral by parts, $ \int_x^1 \frac {\cos t}{t^2}\ dt \equal{} \left[ \minus{} \frac {\cos t}{t}\right]_x^1 \minus{} \int_x^1 \frac {\sin t}{t}\ dt$

$ \equal{} \frac {\cos x}{x} \minus{} \cos 1 \minus{} \int_x^1 \frac {\sin t}{t}\ dt$.

$ \therefore \lim_{x\rightarrow \plus{}0} x\int_x^1 \frac {\cos t}{t^2}\ dt \equal{} \cos 0 \equal{} 1$.

A somewhat different arrangement of the argument:

$ \frac{\cos t}{t^2}\equal{}\frac1{t^2}\plus{}g(t),$ where $ g(t)$ is a bounded function on $ (0,1].$ In fact, $ g(t)$ can be extended to be a continuous function on $ [0,1].$ That means that $ \int_0^1g(t)\,dt$ is some finite constant, that $ \int_x^1g(t)\,dt$ is bounded for $ x\in(0,1],$ and hence $ \lim_{x\to0^\plus{}}x\int_x^1g(t)\,dt\equal{}0.$

Subtract that zero limit from our original problem, and we find that

$ \lim_{x\to0^\plus{}}x\int_x^1\frac{\cos t}{t^2}\,dt\equal{}\lim_{x\to0^\plus{}}x\int_x^1\frac1{t^2}\,dt\equal{}\lim_{x\to0^\plus{}}1\minus{}x\equal{}1.$