Community

Our Precalculus course starts on Dec. 4. Master trig, complex numbers, and vectors and matrices in 2 and 3 dimensions. Click here to enroll today!
Login Register Memberlist Search AoPS Blogs Contests Galleries Forum Index
The time now is Sat Nov 28, 2009 2:23 pm
All times are UTC - 8
View posts since last visit
View unanswered posts
Proof contest
Moderators: Pre-Olympiad Moderators
Post new topic   Reply to topic View previous topicView next topic
193 Posts • Page 2 of 10 • Previous 1, 2, 3, 4, ..., 8, 9, 10 Next
Author Message
chess64
Birch & Swinnerton Dyer
Birch & Swinnerton Dyer


Offline
Joined: 20 Feb 2005
Posts: 4872
PakistanUnited States

To rate posts you must be logged in
#21
Week 4 Solution

The statement of Germany 1996 #2 is the contrapositive of this problem, and a nice solution to that is given here. Some people also used the contrapositive technique for this problem.

PostPosted: Sun May 21, 2006 5:43 am  Back to top 
  ProfilePMBlogAlbum
seesaw
Hodge Conjecture
Hodge Conjecture

Offline
Joined: 09 Mar 2006
Posts: 53
United States

To rate posts you must be logged in
#22
I will take over the proof contest, but I want someone else to help me (as I am not always available.) Are there any volunteers?

Anyway:

Week 5

Prove that, for all values of \theta, (provided that \tan\theta and \sec^{2}\theta are defined), 2\tan\theta \leq \sec^{2}\theta

(PM me your proof)

PostPosted: Sun May 21, 2006 6:41 am  Back to top 
  ProfilePM
darkdevil
P versus NP
P versus NP

Offline
Joined: 21 Sep 2005
Posts: 25
Location: localhost
Romania

To rate posts you must be logged in
#23
now you are the one who gets the solutions? not chess64 anymore?
_________________
\mathfrak{Radvan Marius} aka \mathfrak{darkdevil}

PostPosted: Sun May 21, 2006 10:04 am  Back to top 
  ProfilePMYM
darkdevil
P versus NP
P versus NP

Offline
Joined: 21 Sep 2005
Posts: 25
Location: localhost
Romania

To rate posts you must be logged in
#24
and btw.....what is sec \theta ??
_________________
\mathfrak{Radvan Marius} aka \mathfrak{darkdevil}

PostPosted: Sun May 21, 2006 10:09 am  Back to top 
  ProfilePMYM
seesaw
Hodge Conjecture
Hodge Conjecture

Offline
Joined: 09 Mar 2006
Posts: 53
United States

To rate posts you must be logged in
#25
darkdevil wrote:
now you are the one who gets the solutions? not chess64 anymore?


Yes, I am the one who gets the solutions (for now at least).




darkdevil wrote:
and btw.....what is sec \theta ??


\sec\theta is 1/\cos\theta.

PostPosted: Sun May 21, 2006 10:16 am  Back to top 
  ProfilePM
seesaw
Hodge Conjecture
Hodge Conjecture

Offline
Joined: 09 Mar 2006
Posts: 53
United States

To rate posts you must be logged in
#26
Scoreboard

\begin{tabular}{|l||c|c|c|c|c|c|c|c|}\hline & Week 1 & Week 2 & Week 3 & Week 4 & Week 5 & Week 6 &am...

Last edited by seesaw on Sun Jun 04, 2006 2:51 pm; edited 3 times in total 
PostPosted: Sun May 21, 2006 1:31 pm  Back to top 
  ProfilePM
drunner2007
Navier-Stokes Equations
Navier-Stokes Equations


Offline
Joined: 23 Mar 2005
Posts: 1117
Location: United States
United States

To rate posts you must be logged in
#27
Remark

any new updates/remarks?
_________________
It is never too late to be what you could have been...

The only way to predict the future is to make it.

PostPosted: Thu May 25, 2006 5:33 am  Back to top 
  ProfilePM
seesaw
Hodge Conjecture
Hodge Conjecture

Offline
Joined: 09 Mar 2006
Posts: 53
United States

To rate posts you must be logged in
#28
I might not be available for the next few days, so I am posting week 6 today.

Week 5 Solution
matt276eagles wrote:
Since the maximum of the function f(x)=\sin x is 1, we have \sin 2\theta\le 1 for all values of \theta. Using the double angle identity for sine, we obtain 2\sin\theta\cos\theta\le 1.

Dividing by \cos^{2}\theta, which is positive for all values of \theta such that \tan\theta and \sec^{2}\theta are defined, we have \frac{2\sin\theta}{\cos\theta}\le\frac{1}{\cos^{2}\theta}. By the definitions of tangent and secant, we have 2\tan\theta\le\sec^{2}\theta, which is the desired result.



I would like to remind everybody to try to write proofs that are as clear as possible.

PostPosted: Fri May 26, 2006 11:38 am  Back to top 
  ProfilePM
seesaw
Hodge Conjecture
Hodge Conjecture

Offline
Joined: 09 Mar 2006
Posts: 53
United States

To rate posts you must be logged in
#29
I think that after Week 6 I may start again from week 1 and have proof contests in 6-week cycles. Any feedback on that idea?

Also, does anyone want to volunteer to help out with the proof contest or to be a backup?

Week 6

Let a, b, and c be fixed natural numbers such that, for every positive integer n, there is a triangle whose sides have length a^n, b^n, and c^n. Prove that these triangles are not scalene.

PostPosted: Fri May 26, 2006 11:42 am  Back to top 
  ProfilePM
seesaw
Hodge Conjecture
Hodge Conjecture

Offline
Joined: 09 Mar 2006
Posts: 53
United States

To rate posts you must be logged in
#30
Week 6 Solution
Jiang wrote:
Let a>b>c and a<b+c. For there to be a triangle for every positive integer n, a^n<b^n+c^n must be true for all n. Dividing the inequality by b^n, we get (\frac{a}{b})^n<(\frac{c}{b})^n+1. Because \frac{a}{b}>1, as n gets larger, (\frac{a}{b})^n also gets larger without limit. While because \frac{c}{b}<1, as n gets larger, (\frac{c}{b})^n approaches more and more to 0. So the inequality would eventually prove to be false. If either two or all of the sides are same, it is obviouse that they will always be the same for any n. A scalene triangle must have all of its sides in different lengths, so it would not be possible.


I am sorry to say that due to certain circumstances, I am no longer able to run the Proof Contest. If there are any volunteers to take over the contest please PM me.

PostPosted: Sun Jun 04, 2006 2:54 pm  Back to top 
  ProfilePM
darkdevil
P versus NP
P versus NP

Offline
Joined: 21 Sep 2005
Posts: 25
Location: localhost
Romania

To rate posts you must be logged in
#31
Week 7

f=x^3-x-1
Prove that f has only one real root r \in \left(1, \frac 32 \right)
Prove that if a is a root and a \notin \mathbb{R} then |a| \in \left( \sqrt{\left(\frac 23\right)}, 1\right)
_________________
\mathfrak{Radvan Marius} aka \mathfrak{darkdevil}

PostPosted: Tue Jun 06, 2006 5:48 am  Back to top 
  ProfilePMYM
1=2
Birch & Swinnerton Dyer
Birch & Swinnerton Dyer


Offline
Joined: 06 Jan 2006
Posts: 5825
Location: One Brownie Point!
United States

To rate posts you must be logged in
#32
Since week 7 has no solution posted yet, and it is overdue, I would like to turn this into a proof marathon.

Here's an easy one.

Prove that
3a_{1}+3a_{2}+3a_{3}+\cdot\cdot \cdot+3a_{n}\geq6\sqrt[n]{a_{1}a_{2}a_{3}\cdot \cdot \cdot a_{n}}
_________________
Hmm... apparently quoting Disney movies gets me muted on FTW. Razz

PostPosted: Mon Jul 17, 2006 6:02 am  Back to top 
  ProfilePMWWWBlogAlbum
lotrgreengrapes7926
Navier-Stokes Equations
Navier-Stokes Equations


Offline
Joined: 31 Mar 2006
Posts: 1855
United States

To rate posts you must be logged in
#33
nutz_for2.718281828 wrote:
Since week 7 has no solution posted yet, and it is overdue, I would like to turn this into a proof marathon.

Here's an easy one.

Prove that
3a_{1}+3a_{2}+3a_{3}+\cdot\cdot \cdot+3a_{n}\geq6\sqrt[n]{a_{1}a_{2}a_{3}\cdot \cdot \cdot a_{n}}

Click the smiley. Mr. Green
Huh?
Divide by 3. Then, AM-GM, as long as n\ge2. maybe

_________________
Mr. Green

PostPosted: Mon Jul 17, 2006 6:35 am  Back to top 
  ProfilePM
SplashD
Birch & Swinnerton Dyer
Birch & Swinnerton Dyer

Offline
Joined: 09 Apr 2006
Posts: 2762
Location: Je ne sais pas
ChinaUnited States

To rate posts you must be logged in
#34
Does AM-GM work for more than two numbers?
_________________
"You don't win no silver, you only lose the gold." -David Glen Eisley

PostPosted: Mon Jul 17, 2006 6:37 am  Back to top 
  ProfilePM
lotrgreengrapes7926
Navier-Stokes Equations
Navier-Stokes Equations


Offline
Joined: 31 Mar 2006
Posts: 1855
United States

To rate posts you must be logged in
#35
SplashD wrote:
Does AM-GM work for more than two numbers?
Yes. Mr. Green
OK, this post is not too short anymore.
_________________
Mr. Green

PostPosted: Mon Jul 17, 2006 6:38 am  Back to top 
  ProfilePM
SplashD
Birch & Swinnerton Dyer
Birch & Swinnerton Dyer

Offline
Joined: 09 Apr 2006
Posts: 2762
Location: Je ne sais pas
ChinaUnited States

To rate posts you must be logged in
#36
IF you divide by 3, wouldn't you still have a two on the RHS?
_________________
"You don't win no silver, you only lose the gold." -David Glen Eisley

PostPosted: Mon Jul 17, 2006 6:40 am  Back to top 
  ProfilePM
lotrgreengrapes7926
Navier-Stokes Equations
Navier-Stokes Equations


Offline
Joined: 31 Mar 2006
Posts: 1855
United States

To rate posts you must be logged in
#37
SplashD wrote:
IF you divide by 3, wouldn't you still have a two on the RHS?

OK, I meant divide by 6, to get Click to reveal hidden content
\frac{a_{1}+a_{2}+...+a_{n}}{2}\ge\sqrt[n]{a_{1}a_{2}...a_{n}}.

From AM-GM, \frac{a_{1}+a_{2}+...+a_{n}}{n}\ge\sqrt[n]{a_{1}a_{2}...a_{n}}. Since, \frac{a_{1}+a_{2}+...+a_{n}}{n}\le\frac{a_{1}+a_{2}+...+a_{n}}{2} for all n\ge2, \frac{a_{1}+a_{2}+...+a_{n}}{2}\ge\sqrt[n]{a_{1}a_{2}...a_{n}} must be true.


And nutzfore, you have to state a_{i}>0.
_________________
Mr. Green

PostPosted: Mon Jul 17, 2006 6:44 am  Back to top 
  ProfilePM
matt276eagles
Navier-Stokes Equations
Navier-Stokes Equations


Offline
Joined: 08 Jun 2005
Posts: 1235
Location: Princeton, NJ
United States

To rate posts you must be logged in
#38
nutz_for2.718281828 wrote:
Since week 7 has no solution posted yet, and it is overdue, I would like to turn this into a proof marathon.

Here's an easy one.

Prove that
3a_{1}+3a_{2}+3a_{3}+\cdot\cdot \cdot+3a_{n}\geq6\sqrt[n]{a_{1}a_{2}a_{3}\cdot \cdot \cdot a_{n}}


Huh? For n=1 and a_{1}=1, the inequality gives 3\ge 6.

PostPosted: Mon Jul 17, 2006 9:53 am  Back to top 
  ProfilePM
SplashD
Birch & Swinnerton Dyer
Birch & Swinnerton Dyer

Offline
Joined: 09 Apr 2006
Posts: 2762
Location: Je ne sais pas
ChinaUnited States

To rate posts you must be logged in
#39
matt276eagles wrote:
nutz_for2.718281828 wrote:
Since week 7 has no solution posted yet, and it is overdue, I would like to turn this into a proof marathon.

Here's an easy one.

Prove that
3a_{1}+3a_{2}+3a_{3}+\cdot\cdot \cdot+3a_{n}\geq6\sqrt[n]{a_{1}a_{2}a_{3}\cdot \cdot \cdot a_{n}}


Huh? For n=1 and a_{1}=1, the inequality gives 3\ge 6.


lotrgreengrapes said n \ge 2
_________________
"You don't win no silver, you only lose the gold." -David Glen Eisley

PostPosted: Mon Jul 17, 2006 9:58 am  Back to top 
  ProfilePM
amcavoy
Yang-Mills Theory
Yang-Mills Theory

Offline
Joined: 21 Mar 2005
Posts: 922
United States

To rate posts you must be logged in
#40
Here's another easy one:

For n=3, prove the AM-GM inequality. That is, prove that \frac{a+b+c}{3}\geq\sqrt[3]{abc}.

PostPosted: Mon Jul 17, 2006 10:24 am  Back to top 
  ProfilePM
Display posts from previous:   Sort by:   
193 Posts • Page 2 of 10 • Previous 1, 2, 3, 4, ..., 8, 9, 10 Next
Post new topic   Reply to topic View previous topicView next topic
Jump to:  

You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
You cannot attach files in this forum
You can download files in this forum
You cannot post calendar events in this forum


© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us