geometry
geometric transformation
reflection
HCSSiM

All right, we are starting a new game. But before we do, I have some rule change proposals:

Proposal A: Once a biohazard suit is bought, it is permanent.
Proposal B: Shields will protect against conventional attacks for the same duration, and can prevent the first missile attack that occurs in the duration.
Proposal C: Each state start with 35 points.
Proposal D: Each state has a maximum of 1 player.
Proposal E: Nuclear Defense Systems and SAM Rocket-Intercepts cost 35 instead of 50.
Proposal F: Nukes cost 40.

Here is most of my reasoning.

[quote="chenhsi"]Anyway, some general comments on the game:

Shields are pointless: Conventional attacks are almost never used, and missiles are used instead. Maybe make the shield protect against rocket attacks as well, but only prevent a limited amount of damage?

Biohazard Suits: Too expensive. Spending 25 points to protect against 5 damage? This should be changed so that once you buy the suit, you can NEVER get any nuke splash damage. This is still balance, since nukes are so rarely used.

Nuke Defense + Rocket Intercept: Rarely used. Consider making cheaper.

Nukes: Rarely used, since V2s can take out an undefended state, and a maxxed out alliance can do 9 damage with conventional attacks, for a total of 34 damage.

My recommendation: Start with more points, maybe 40. It is too easy to destroy an state with no players. 1 V2 eliminates a state. You saw how I used the combined forces of my alliance to destroy 6 states in 1 turn? Yeah.

Also, why don't you make it 1 person/state max? There isn't that many players. This way, we also don't have to check whether there are 3 players in a state, we only need to check for 1.[/quote]

A, accept
B, partial accept. Once a missile hits a shield the shield should dissipate.
C, accept
D, accept
E, cost should be 40.
F, reject: nukes KILL a state instantly + possibly 25-40 extra splash damage.

those are my votes, waiting for others to check proposal...

New Proposals:

Proposal G: You may only buy one item per post [e.g. no buying 3 V2 Rockets to blow up states in one post]
Proposal H: 1-hour post limit, no double-posting
Proposal I: V2 cost = 25, V1 cost = 12
Proposal J: Kills give 3 pts to killer

B. Yes, I did mean that when a missile hits a shield the shield disappears. Sorry if I wasn't obvious.
E. Sure.
F. I meant that it should cost 40 to buy a nuke. How about 40 to buy, kills a state instantly and does 10 splash damage? 25-40 would probably kill any nearby states as well. 10 would severly weaken.

G. No.
H. No. I disagree with any rule that discourages being on AoPS more.
I. Sure.
J. Sure.

Edit: Proposal K: The game wins when only 1 alliance is left standing. If the alliance wants to disband and keep fighting, fine, but the states are not forced to kill each other.

How about adding mini-rockets?  So that...

V1 Rocket- 5 cost, 5 dmg
V2 Rocket- 10 cost, 10 dmg
V3 Rocket- 12 cost, 15 dmg
V4 Rocket- 15 cost, 20 dmg
V5 Rocket- 20 cost, 25 dmg
V10 Rocket- 40 cost, 50 dmg

Mini-Nuke- 30 cost, 5 dmg everytime you post within 10 hours
Nuke- 50 cost, automatically kills a state, -10 to states surrounding the nuked state
Ultra-Nuke- 100 cost, automatically kills a state, kills the states surrounding that state

Shield- 5 cost, protects you from 1 point damage attack
Strong Shield- 8 cost, protects you from 2 point damage attacks
Day Shield- 10 cost, protects you from 3 point damage attacks
Rocket Shield- 15 cost, protects you from 1 rocket attack
Strong Rocket Shield- 25 cost, protects you from 2 rocket attacks
Day Rocket Shield- 40 cost, protects you from 3 rocket attacks
Biohazard Defense System- 60 cost, protects you from nukes
Rocket Stealer- 60 cost, when you are attacked by a rocket, take that rocket and fire it back at the state that fired that rocket.  works twice.

yay typing.

A: accept
B: accept
C: accept
D: neutral
E: Accept
F: neutral
G: reject; if someone wants to do that, we can all gang up on him/her or his/her alliance and destroy
H: partial reject; how about half-hour limits?
I: neutral
J: accept
K: accept
Ernie: mini rockets??? no two versions is enough. mini nuke is bad because someone can dish out 100 dmg max. Ultra nuke is a bit powerful, but it does come with a heavy cost. The only drawback is if there is a four person alliance with all 100 health, they can split the cost and kill 8 states max. They can do it two more times if they are crazy and G is rejected. Your shields, um some of them might not be rejected, but if a state has lots of hp, it can buy a biohazard defense system and a rocket stealer and pwn everybody if there is no expiration date.

A: 3 accepts
B: 3 accepts
C: 3 accepts
D: neutral, 2 accepts
E: 3 accepts
F: neutral, accept
G: 2 rejects, accept
H: partial reject, full reject, accept
I: neutral, 2 accepts
J: 3 accepts
K: 2 accepts
Ernie's Items: 2 Rejects

James, you need to vote on Item F (I know you rejected it, but I think you misunderstood the item), Item K, and Ernie's Items.

Ernie, you need to vote on all of them.

I would like it if stevenmeow voted, since he was in the other game.

Partial reject to Prop F. 10 pts change for nuke price is not going to change the outcome of the war all that much.

On ernie's new items: normal shields are fine, rocket shields don't look fine, stealer is terrible. Mini-nuke: Don't get it. Nuke: Accept. Ultra-Nuke: Reject. WAY too powerful. V3-10 Rockets :too repetitive. Two versions are enough. Also if Prop G is axed out, then you could just "stack" multiple rockets to form one big rocket.

[Ernie's Items final proposal: partial reject]

Proposal K: Neutral. Seriously I don't care if they want to keep going or not.

New Proposal L: Biohazard suits: They should ameliorate splash damage, not eliminate it. [Say maybe lower splash damage from 10 to 2?]

This is to reflect the fact that Biohazard suits don't 100% block out radiation.

1 hr post limit => Weak Accept

[code]
Prop.A..a..N..r..R..Description
A  3 - - - - Permanent Biohazard Suits
B  3 - - - - Missile-Preventing Shields
C  3 - - - - 35pts start
D  2 - 1 - - 1 player states
E  3 - - - - 35pt cost for Rocket/Nuke defense
F  1 - 1 1 - 40pt Nukes
G  1 - - - 2 1 item per post
H  - 1 - 1 1 1hr post limit
I  2 - 1 - - 12/25 cost for V1/V2 rockets
J  3 - - - - Kills give 3pts to killer
K  2 - 1 - - Game ends at single alliance
L  1 - - - - [NEW] Weaker Biohazard Suits
E.I - - - 1 2 Addition of new items

Note: E.I = Ernie's Items
A = Accept, a = Weak/Partial Accept, N = Neutral, r = Weak/Partial Reject, R = Reject

[/code]

[code]
Prop.A..a..N..r..R..Description
A  3 - - - - Permanent Biohazard Suits
B  3 - - - - Missile-Preventing Shields
C  3 - - - - 35pts start
D  2 - 1 - - 1 player states
E  3 - - - - 35pt cost for Rocket/Nuke defense
F  1 - 1 1 - 40pt Nukes
G  1 - - - 2 1 item per post
H  - 1 - 1 1 1hr post limit
I  2 - 1 - - 12/25 cost for V1/V2 rockets
J  3 - - - - Kills give 3pts to killer
K  2 - 1 - - Game ends at single alliance
L  1 - - 1 - [NEW] Weaker Biohazard Suits
E.I - - - 1 2 Addition of new items

Note: E.I = Ernie's Items
A = Accept, a = Weak/Partial Accept, N = Neutral, r = Weak/Partial Reject, R = Reject

[/code]

Erm, when is the Civil war actually going to take place?

The states are at peace right now, trying to figure out what weapons to allow and what weapons not to allow.

 :P

If chenhsi and ernie both live in CA, it would be more like state. In reality, states just attack when their opponent is still arguing about which items to allow.

+4 Califorina, -5 Georgia

[hide="states"][color=blue]Alabama-39 pts
Alaska-40 pts
California-42 pts
Colorado-55 pts
Connecticut-39 pts[/color]
[color=green]Delaware-25 pts
Florida-27 pts
Georgia-30 pts
Hawaii-35 pts
Idaho-35 pts
Illinois-35 pts
Indiana-35 pts
Iowa-35 pts
Kansas-35 pts
Kentucky-35 pts
Louisiana-35 pts
Maine-35 pts
Maryland-35 pts
Massachusetts-35 pts
Michigan-35 pts
Minnesota-35 pts
Mississippi-35 pts
Missouri-43 pts
Montana-35 pts
Nebraska-35 pts
Nevada-35 pts
New Hampshire-35 pts
New Jersey-35 pts
New Mexico-35 pts
New York-35 pts
North Carolina-35 pts
North Dakota-35 pts
Ohio-35 pts
Oklahoma-35 pts
Oregon-35 pts
Pennsylvania-35 pts
Rhode Island-35 pts
South Carolina-35 pts
South Dakota-35 pts
Tennessee-35 pts
Texas-35 pts
Utah-35 pts
Vermont-35 pts
Virginia-35 pts
Washington-35 pts
West Virginia-35 pts
Wisconsin-35 pts
Wyoming-35 pts[/color][/hide][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote]

+5pts alaska, -4 delaware

[hide="states"][color=blue]Alabama-39 pts
Alaska-45 pts
California-42 pts
Colorado-55 pts
Connecticut-39 pts[/color]
[color=green]Delaware-21 pts
Florida-27 pts
Georgia-30 pts
Hawaii-35 pts
Idaho-35 pts
Illinois-35 pts
Indiana-35 pts
Iowa-35 pts
Kansas-35 pts
Kentucky-35 pts
Louisiana-35 pts
Maine-35 pts
Maryland-35 pts
Massachusetts-35 pts
Michigan-35 pts
Minnesota-35 pts
Mississippi-35 pts
Missouri-43 pts
Montana-35 pts
Nebraska-35 pts
Nevada-35 pts
New Hampshire-35 pts
New Jersey-35 pts
New Mexico-35 pts
New York-35 pts
North Carolina-35 pts
North Dakota-35 pts
Ohio-35 pts
Oklahoma-35 pts
Oregon-35 pts
Pennsylvania-35 pts
Rhode Island-35 pts
South Carolina-35 pts
South Dakota-35 pts
Tennessee-35 pts
Texas-35 pts
Utah-35 pts
Vermont-35 pts
Virginia-35 pts
Washington-35 pts
West Virginia-35 pts
Wisconsin-35 pts
Wyoming-35 pts[/color][/hide][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote]

-9 Delaware

[hide="states"][color=blue]Alabama-39 pts
Alaska-45 pts
California-42 pts
Colorado-55 pts
Connecticut-39 pts[/color]
[color=green]Delaware-12 pts
Florida-27 pts
Georgia-30 pts
Hawaii-35 pts
Idaho-35 pts
Illinois-35 pts
Indiana-35 pts
Iowa-35 pts
Kansas-35 pts
Kentucky-35 pts
Louisiana-35 pts
Maine-35 pts
Maryland-35 pts
Massachusetts-35 pts
Michigan-35 pts
Minnesota-35 pts
Mississippi-35 pts
Missouri-43 pts
Montana-35 pts
Nebraska-35 pts
Nevada-35 pts
New Hampshire-35 pts
New Jersey-35 pts
New Mexico-35 pts
New York-35 pts
North Carolina-35 pts
North Dakota-35 pts
Ohio-35 pts
Oklahoma-35 pts
Oregon-35 pts
Pennsylvania-35 pts
Rhode Island-35 pts
South Carolina-35 pts
South Dakota-35 pts
Tennessee-35 pts
Texas-35 pts
Utah-35 pts
Vermont-35 pts
Virginia-35 pts
Washington-35 pts
West Virginia-35 pts
Wisconsin-35 pts
Wyoming-35 pts[/color][/hide][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote]

+9 California

[hide="states"][color=blue]Alabama-39 pts
Alaska-45 pts
California-51 pts
Colorado-55 pts
Connecticut-39 pts[/color]
[color=green]Delaware-12 pts
Florida-27 pts
Georgia-30 pts
Hawaii-35 pts
Idaho-35 pts
Illinois-35 pts
Indiana-35 pts
Iowa-35 pts
Kansas-35 pts
Kentucky-35 pts
Louisiana-35 pts
Maine-35 pts
Maryland-35 pts
Massachusetts-35 pts
Michigan-35 pts
Minnesota-35 pts
Mississippi-35 pts
Missouri-43 pts
Montana-35 pts
Nebraska-35 pts
Nevada-35 pts
New Hampshire-35 pts
New Jersey-35 pts
New Mexico-35 pts
New York-35 pts
North Carolina-35 pts
North Dakota-35 pts
Ohio-35 pts
Oklahoma-35 pts
Oregon-35 pts
Pennsylvania-35 pts
Rhode Island-35 pts
South Carolina-35 pts
South Dakota-35 pts
Tennessee-35 pts
Texas-35 pts
Utah-35 pts
Vermont-35 pts
Virginia-35 pts
Washington-35 pts
West Virginia-35 pts
Wisconsin-35 pts
Wyoming-35 pts[/color][/hide][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote]

-2 Delaware, -7 Florida

[hide="states"][color=blue]Alabama-39 pts
Alaska-45 pts
California-51 pts
Colorado-55 pts
Connecticut-39 pts[/color]
[color=green]Delaware-10 pts
Florida-20 pts
Georgia-30 pts
Hawaii-35 pts
Idaho-35 pts
Illinois-35 pts
Indiana-35 pts
Iowa-35 pts
Kansas-35 pts
Kentucky-35 pts
Louisiana-35 pts
Maine-35 pts
Maryland-35 pts
Massachusetts-35 pts
Michigan-35 pts
Minnesota-35 pts
Mississippi-35 pts
Missouri-43 pts
Montana-35 pts
Nebraska-35 pts
Nevada-35 pts
New Hampshire-35 pts
New Jersey-35 pts
New Mexico-35 pts
New York-35 pts
North Carolina-35 pts
North Dakota-35 pts
Ohio-35 pts
Oklahoma-35 pts
Oregon-35 pts
Pennsylvania-35 pts
Rhode Island-35 pts
South Carolina-35 pts
South Dakota-35 pts
Tennessee-35 pts
Texas-35 pts
Utah-35 pts
Vermont-35 pts
Virginia-35 pts
Washington-35 pts
West Virginia-35 pts
Wisconsin-35 pts
Wyoming-35 pts[/color][/hide][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote]

-9 Missouri

[hide="states"][color=blue]Alabama-39 pts
Alaska-45 pts
California-51 pts
Colorado-55 pts
Connecticut-39 pts[/color]
[color=green]Delaware-10 pts
Florida-20 pts
Georgia-30 pts
Hawaii-35 pts
Idaho-35 pts
Illinois-35 pts
Indiana-35 pts
Iowa-35 pts
Kansas-35 pts
Kentucky-35 pts
Louisiana-35 pts
Maine-35 pts
Maryland-35 pts
Massachusetts-35 pts
Michigan-35 pts
Minnesota-35 pts
Mississippi-35 pts[/color]
Missouri-34 pts
[color=green]Montana-35 pts
Nebraska-35 pts
Nevada-35 pts
New Hampshire-35 pts
New Jersey-35 pts
New Mexico-35 pts
New York-35 pts
North Carolina-35 pts
North Dakota-35 pts
Ohio-35 pts
Oklahoma-35 pts
Oregon-35 pts
Pennsylvania-35 pts
Rhode Island-35 pts
South Carolina-35 pts
South Dakota-35 pts
Tennessee-35 pts
Texas-35 pts
Utah-35 pts
Vermont-35 pts
Virginia-35 pts
Washington-35 pts
West Virginia-35 pts
Wisconsin-35 pts
Wyoming-35 pts[/color][/hide][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote]

+9 California

[hide="states"][color=blue]Alabama-39 pts
Alaska-45 pts[/color]
[color=orange]California-60 pts[/color]
[color=blue]Colorado-55 pts
Connecticut-39 pts[/color]
[color=green]Delaware-10 pts
Florida-20 pts
Georgia-30 pts
Hawaii-35 pts
Idaho-35 pts
Illinois-35 pts
Indiana-35 pts
Iowa-35 pts
Kansas-35 pts
Kentucky-35 pts
Louisiana-35 pts
Maine-35 pts
Maryland-35 pts
Massachusetts-35 pts
Michigan-35 pts
Minnesota-35 pts
Mississippi-35 pts
Missouri-34 pts
Montana-35 pts
Nebraska-35 pts
Nevada-35 pts
New Hampshire-35 pts
New Jersey-35 pts
New Mexico-35 pts
New York-35 pts
North Carolina-35 pts
North Dakota-35 pts
Ohio-35 pts
Oklahoma-35 pts
Oregon-35 pts
Pennsylvania-35 pts
Rhode Island-35 pts
South Carolina-35 pts
South Dakota-35 pts
Tennessee-35 pts
Texas-35 pts
Utah-35 pts
Vermont-35 pts
Virginia-35 pts
Washington-35 pts
West Virginia-35 pts
Wisconsin-35 pts
Wyoming-35 pts[/color][/hide][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote]

Pythag already staked CA. pick another state

Alaska: +9

then V2 delaware, florida

[hide="states"][color=blue]Alabama-41 pts
Alaska-10 pts
California-62 pts
Colorado-57 pts
Connecticut-41 pts[/color]
[color=green]
Georgia-30 pts
Hawaii-35 pts
Idaho-35 pts
Illinois-35 pts
Indiana-35 pts
Iowa-35 pts
Kansas-35 pts
Kentucky-35 pts
Louisiana-35 pts
Maine-35 pts
Maryland-35 pts
Massachusetts-35 pts
Michigan-35 pts
Minnesota-35 pts
Mississippi-35 pts
Missouri-34 pts
Montana-35 pts
Nebraska-35 pts
Nevada-35 pts
New Hampshire-35 pts
New Jersey-35 pts
New Mexico-35 pts
New York-35 pts
North Carolina-35 pts
North Dakota-35 pts
Ohio-35 pts
Oklahoma-35 pts
Oregon-35 pts
Pennsylvania-35 pts
Rhode Island-35 pts
South Carolina-35 pts
South Dakota-35 pts
Tennessee-35 pts
Texas-35 pts
Utah-35 pts
Vermont-35 pts
Virginia-35 pts
Washington-35 pts
West Virginia-35 pts
Wisconsin-35 pts
Wyoming-35 pts[/color][/hide]

+3 GA
Buys V1, destroys AK, +3 GA for the kill.
Casualties: "13 people and some caribou."
[hide="states"]
[color=blue]Alabama-41 pts[/color]
Alaska-PWNT
[color=blue]California-62 pts[/color]
[color=blue]Colorado-57 pts[/color]
[color=blue]Connecticut-41 pts[/color]
Delaware-PWNT
Florida-PWNT
[color=green]Georgia-26 pts
Hawaii-35 pts
Idaho-35 pts
Illinois-35 pts
Indiana-35 pts
Iowa-35 pts
Kansas-35 pts
Kentucky-35 pts
Louisiana-35 pts
Maine-35 pts
Maryland-35 pts
Massachusetts-35 pts
Michigan-35 pts
Minnesota-35 pts
Mississippi-35 pts
Missouri-34 pts
Montana-35 pts
Nebraska-35 pts
Nevada-35 pts
New Hampshire-35 pts
New Jersey-35 pts
New Mexico-35 pts
New York-35 pts
North Carolina-35 pts
North Dakota-35 pts
Ohio-35 pts
Oklahoma-35 pts
Oregon-35 pts
Pennsylvania-35 pts
Rhode Island-35 pts
South Carolina-35 pts
South Dakota-35 pts
Tennessee-35 pts
Texas-35 pts
Utah-35 pts
Vermont-35 pts
Virginia-35 pts
Washington-35 pts
West Virginia-35 pts
Wisconsin-35 pts
Wyoming-35 pts[/color][/hide]

...

Why did you just kill our ally???

-7 Georgia

[hide="states"]
[color=blue]Alabama-41 pts[/color]
Alaska-PWNT
[color=blue]California-62 pts[/color]
[color=blue]Colorado-57 pts[/color]
[color=blue]Connecticut-41 pts[/color]
Delaware-PWNT
Florida-PWNT
[color=green]Georgia-19 pts
Hawaii-35 pts
Idaho-35 pts
Illinois-35 pts
Indiana-35 pts
Iowa-35 pts
Kansas-35 pts
Kentucky-35 pts
Louisiana-35 pts
Maine-35 pts
Maryland-35 pts
Massachusetts-35 pts
Michigan-35 pts
Minnesota-35 pts
Mississippi-35 pts
Missouri-34 pts
Montana-35 pts
Nebraska-35 pts
Nevada-35 pts
New Hampshire-35 pts
New Jersey-35 pts
New Mexico-35 pts
New York-35 pts
North Carolina-35 pts
North Dakota-35 pts
Ohio-35 pts
Oklahoma-35 pts
Oregon-35 pts
Pennsylvania-35 pts
Rhode Island-35 pts
South Carolina-35 pts
South Dakota-35 pts
Tennessee-35 pts
Texas-35 pts
Utah-35 pts
Vermont-35 pts
Virginia-35 pts
Washington-35 pts
West Virginia-35 pts
Wisconsin-35 pts
Wyoming-35 pts[/color][/hide][/quote]

@Xantos: Sarah Palin will NOT be happy about this...

Note: Alaka not dead.

james4l made a typo. I don't think Alaska had 10 points at the time: they had 45.

-7 Georgia

[hide="states"]
[color=blue]Alabama-41 pts[/color]
Alaska-PWNT
[color=blue]California-62 pts[/color]
[color=blue]Colorado-57 pts[/color]
[color=blue]Connecticut-41 pts[/color]
Delaware-PWNT
Florida-PWNT
[color=green]Georgia-12 pts
Hawaii-35 pts
Idaho-35 pts
Illinois-35 pts
Indiana-35 pts
Iowa-35 pts
Kansas-35 pts
Kentucky-35 pts
Louisiana-35 pts
Maine-35 pts
Maryland-35 pts
Massachusetts-35 pts
Michigan-35 pts
Minnesota-35 pts
Mississippi-35 pts
Missouri-34 pts
Montana-35 pts
Nebraska-35 pts
Nevada-35 pts
New Hampshire-35 pts
New Jersey-35 pts
New Mexico-35 pts
New York-35 pts
North Carolina-35 pts
North Dakota-35 pts
Ohio-35 pts
Oklahoma-35 pts
Oregon-35 pts
Pennsylvania-35 pts
Rhode Island-35 pts
South Carolina-35 pts
South Dakota-35 pts
Tennessee-35 pts
Texas-35 pts
Utah-35 pts
Vermont-35 pts
Virginia-35 pts
Washington-35 pts
West Virginia-35 pts
Wisconsin-35 pts
Wyoming-35 pts[/color][/hide][/quote][/quote]

Now we only get 7 points per turn :(

I hate you Xantos...temporarily, anyway

-2 Georgia, +5 Colorado, buy V1 rocket, fire at Georgia, gain points for kill

[hide="states"]
[color=blue]Alabama-42 pts[/color]
[color=blue]California-63 pts[/color]
[color=blue]Colorado-53 pts[/color]
[color=blue]Connecticut-42 pts[/color]
[color=green]
Hawaii-35 pts
Idaho-35 pts
Illinois-35 pts
Indiana-35 pts
Iowa-35 pts
Kansas-35 pts
Kentucky-35 pts
Louisiana-35 pts
Maine-35 pts
Maryland-35 pts
Massachusetts-35 pts
Michigan-35 pts
Minnesota-35 pts
Mississippi-35 pts
Missouri-34 pts
Montana-35 pts
Nebraska-35 pts
Nevada-35 pts
New Hampshire-35 pts
New Jersey-35 pts
New Mexico-35 pts
New York-35 pts
North Carolina-35 pts
North Dakota-35 pts
Ohio-35 pts
Oklahoma-35 pts
Oregon-35 pts
Pennsylvania-35 pts
Rhode Island-35 pts
South Carolina-35 pts
South Dakota-35 pts
Tennessee-35 pts
Texas-35 pts
Utah-35 pts
Vermont-35 pts
Virginia-35 pts
Washington-35 pts
West Virginia-35 pts
Wisconsin-35 pts
Wyoming-35 pts[/color][/hide][/quote][/quote][/quote]


[i]Author of the problem is Adrian Satja Kurdija (although, of course, it may be possible that there already exists a very similar or same problem author didn't know about). This was Problem 3 (Geometry) for 9th grade on 1st Balkan Student Mathematical Competition 2008 - a competition 11 schools from Balkan area took part of during November 2008. All information about the competition can be found on http://www.mathlinks.ro/viewtopic.php?t=243614.[/i]

An acute-angled triangle $ ABC$ is given. Bisector of angle $ \measuredangle{BAC}$ intersects the side $ \overline{BC}$ in point $ K$. Point $ L$ is the midpoint of segment $ \overline{AK}$ and point $ N$ is the intersection of altitude from vertex $ A$ and side $ \overline{BC}$. On side $ \overline{AB}$, point $ P$ is such that $ \measuredangle{KLP} \equal{} \measuredangle{BAC}$. Circumcircle of triangle $ ANP$ intersects side $ \overline{AC}$ in point $ Q$. Prove that $ |AP| \equal{} |AQ|$.

Because $ \angle{PAL}\equal{}\frac{\angle{BAC}}{2}\equal{}\frac{\angle{PLK}}{2}$, triangle $ PLA$ is isosceles and $ AL\equal{}PL$. But $ AL\equal{}LN\equal{}LK$, so $ L$ is circumcenter of $ \triangle ANP$ and $ AK$ is diameter of this circle. Therefore $ \angle{KPA}\equal{}\angle{ADK}\equal{}90^\circ$, but $ KP\equal{}KD$, thus $ AP\equal{}AD$.

Note that $ \angle PLK\equal{}\angle APL\plus{}\angle PAL$, i.e., $ \angle APL\equal{}\angle PLK\minus{}\angle PAL\equal{}\angle BAC\minus{}\angle PAL\equal{}\angle LAC\equal{}\angle PAL$. So $ \angle APL\equal{}\angle PAL$ and hence $ AL\equal{}PL$.

Consider the circumcircle of triangle $ ANK$. Since $ \angle ANK\equal{}90^{\circ}$, by Thales' Theorem, $ L$ is the center of the circle, and hence $ LK\equal{}LN\equal{}LA$. So $ LA\equal{}LP\equal{}LN\equal{}LK$. Thus, the circumcircle of triangle $ APN$ coincides with the circumcircle of triangle $ ANK$. Now we have $ LP\equal{}LQ$. Consider triangle $ APL$ and triangle $ ALQ$. Since $ LP\equal{}LQ,LA\equal{}LA,\angle PAL\equal{}\angle LAQ$, they are congruent, which implies $ AP\equal{}AQ$.


[quote="Interval"]How do I factor P^4+5P^2+6?[/quote]

[hide]Assume that $P^{2}=a$, so now we have $a^{2}+5a+6$.

So this can be written as $(a+3)(a+2)$.

Substituting, we get $(P^{2}+3)(P^{2}+2)$[/hide]

I like the way you factored that question.

Thanks, but that was pretty much the only way you could do it.


The poll should be answered as a fraction of the height between two consecutive lines on notebook paper (the US kind that has 33 (not 26) lines).

Also answer these questions if you feel like it:
-Do people sometimes look at your paper and think it is blank when you have actually written all over it?
-Does your English teacher assign a 2 page written essay to everyone, except you, who are assigned a 5 page essay?
-What is the width of your letter "b"s as a fraction of the line height?
-Do you write in the margin?
-Can you normally distinguish between normal and italics while writing?
-How sharp is your pencil, or do you prefer mechanical?
-Do you write your lower case "a"s like alphas (computer font) or the other o-with-a-tail way?
-Do you cross any of your 7's, 0's, z's, or q's?
-What is the height of your lower case "a"s as a fraction of the height of your letter "b"s?
-How dark do you write?

I write very very skinny laterally, but generously heightwise. I'd say 2/3 the width of most people. I think the reason is because I write cursive, and I like to make my curves very elliptical. Sometimes they turn sharp which doesn't look good, but whatever, it's fast.

well..
If i'm in a rush, i stretch out all my letters and it looks like scribbles.

teehee

if i'm writing nicely, my b's are pretty tall and my handwriting is fat.

i make my a's not the computer way.
i don't cross any 7s, 0s, zs, or qs.
my lowercase a's are about 2/3 of my b's.
and i press pretty firmly most of the time, so it's pretty dark.

Yay! People responded!

My b's have a 5:3 height:width ratio :lol: . The b's height is 1/2 the line height (most people think this is small). The darkness of my writing is inversely proportional to the sharpness of the pencil. I write medium-dark with a pen. I cross my 7's, z's and q's.

Another question: Dot i first or write i first? I write the i first.

My handwriting is okay. Sometimes my pencil can get pretty dull, but I try to sharpen them regularly. Ideally, my pencil will be somewhat sharp, but not fresh out of the sharpener. I cross my z's, but nothing else. I stay pretty much in the margins.

argh i hate my handwriting. my teachers can read it barley, but on serious stuff like USAMO i have to spend SO MUCH TIME trying to get it right. I even had to draw lines to not mess up...

[quote="tjhance"]I stay pretty much in the margins.[/quote]

Really? Wow.  :wink:

I fit the size of my writing to the space between lines on the page.  But generally, a "b" takes up about 7/8 of the line height.

When I'm not in a rush, my handwriting is really neat.  But when I am, say, for note taking, it's pretty messy.

I like writing with pens.  If I have to use a pencil, I really don't care what kind I use.
I write my lowercase a's in the "O-with-a-tail" style.
I cross my z's and sometimes I put a "tail" on my q's.  I don't cross my 7's or 0's, but in my neat, not-in-a-rush handwriting, my 7's have a "tail" in the top-left, like in Roman font.

When it's not lined paper (like the USAMO), I tend to write in the size I think will fit my desperate attempt at a solution.  I tend not to draw lines in case I have to draw a diagram.  Lined paper can sometimes get in the way of diagrams, depending on how dark the lines are.

I am about 2/3
but when I write fast I am about 4/5

Wow, nice little bell curve looking thingy going on there.

[quote="moogra"][quote="K81o7"]I think some of those poll options aren't going to come up.  Seriously, who writes b's 1/8 of the height of the space between the lines?[/quote]

Like 5/4 as well. I don't think anyone goes above the lines on every single letter.[/quote]

Really? It's strange that there aren't many extremes to me. There are two people at my school who write really small. I would say their writing is closer to 1/8 than 1/4. The title/heading of their papers is about 1/4... Both of them are very, very, consistent in their handwriting size.

Then there's that girl in my Spanish class who has permission from the teacher to write her answers on a separate sheet of notebook paper instead of the worksheets... I'd say her writing is at a 2.5.

[quote="leoxnlin"]Seems like no one gets what I mean by crossing your q's.... my Spanish teacher is the only person I know who does it, and I got it from her.[/quote]

I cross my q's.

I cross the 7 and put the upper tail, too. My b is about 7/8 or 4/5 (normally. If I want to make a good writing, about 3/5  :rotfl: and if writing with hurry, 5/4). I NEVER cross the 0's, except if I need to write a capitalized O next to a 0. I hate margins, and I hate the test sheets where the line of "email" is as short as date line. IT SUCKS!

[quote="pacman2812"]
I cross the 7 and put the upper tail, too.[/quote]

Do you give your 1's an upper tail? A baseline?

my handwriting is fairly even, not too tall, not too fat, i write my g's kind of like those in times new roman font (two loops), and i cross my z's, 7's, q's, and occasionally (rarely) my 0's.

when i'm very rushed, i write in spikey italics. oh. and i never capitalize (unless i'm taking an english test, in which case i still don't capitalize my name)


Let $a,b,c>1$.Prove that $(log_b a+log_c a).log_a (\frac{b+c}{2})\geq2$ ;)

This problem seems very cute.

By AM-GM, $log_a(\frac{b+c}{2}) \geq log_a(\sqrt{bc}) = \frac{1}{2}(log_a(b)+log_a(c)).$

Thus, $(log_b(a)+log_c(a))log_a(\frac{b+c}{2}) \geq \frac{1}{2}(log_b(a)+log_c(a))(log_a(b)+log_a(c)) = $
$\frac{1}{2}(log_b(a)*log_a(b)+log_b(a)*log_a(c)+log_c(a)*log_a(b)+log_c(a)*log_a(c)).$
By log rules, we can simplify this to $1+\frac{1}{2}(log_b(c)+log_c(b)).$ By a final application of AM-GM, this quantity is greater than or equal to $1+\sqrt{log_b(c)*log_c(b)} = 1+1 = 2.$ In conclusion, $(log_b(a)+log_c(a))log_a(\frac{b+c}{2}) \geq 2.$ QED


linear algebra
matrix
complex numbers
real analysis
real analysis unsolved

I am stuck with something that must be a known result that I have forgotten..

Is it true that given a_i,b_i (complex in general)

Sum 1_N a_i*exp(b_i*n) = Sum 1_M a'_j*exp(b'_j*n)

for n = 0,1,..2*M-1 (so that we have enough equations to compute a'_j, b'_j)

=> (a'_j,b'_j) = (a_i,b_i) ie the exponents and coefficient need to be the same
and so this is possible only if M>=N and if M>N some (a'_j,b'_j) =(0,0)

If so how to prove this?
Please help
Nick

You should be careful here: adding $2\pi i$ to any of $b_j$ won't change the value of the exponent at any integer point. Also, you probably assume that $M\ge N$ (otherwise you have more variables $a_j, a'_j$ in your linear system than equations).  Also, of course, you have stupid identities like $2e^n=e^n+e^n$.
So, let us restate your question as follows:

Let $a_k$, $b_k$ ($1\le j\le K$) be complex numbers, let $b_j$ be all different and satisfy $-\pi\le\Im b_j<\pi$. Is it true that the conditions $\sum_{j=1}^K a_j e^{b_j n}=0$ ($0\le n\le K-1$) imply that all $a_j=0$?

Then the  answer is yes. All you need to do is to look at these equations as at a linear system of $K$ equations with $K$ unknowns and recall how to compute the determinant of the matrix of the system (the Vandermonde determinant).

Sorry for the double post and thnx for the answer.It was clear but I had posted it before this one and forgot about that.My excuses.Ok??
Nick


geometry
geometric transformation
rotation
linear algebra
matrix
induction

If $ A_{(\theta)}$ is a rotational matrix, show that for angles $ \alpha$ and $ \beta$ we have

$ A_{(\alpha)}\cdot A_{(\beta)}\equal{}A_{(\alpha\plus{}\beta)}$

[quote="Kent Merryfield"]You must mean to restrict this question to $ 2\times 2$ matrices, right? Otherwise, it wouldn't be clear what $ \alpha,$ $ \beta,$ or $ \alpha \plus{} \beta$ would mean.[/quote]

Yes, didn't think I needed to mention this, sorry.

[quote="QuyBac"][hide="Hint"]$ A_{(\theta)} \equal{} \begin{bmatrix}cos(\theta) & \minus{} sin(\theta) \\
sin(\theta) & cos(\theta)\end{bmatrix}$
 $ A_{(\alpha)}\cdot A_{(\beta)} \equal{} \begin{bmatrix}cos(\alpha) & \minus{} sin(\alpha) \\
sin(\alpha) & cos(\alpha)\end{bmatrix}\begin{bmatrix}cos(\beta) & \minus{} sin(\beta) \\
sin(\beta) & cos(\beta)\end{bmatrix} \equal{} \begin{bmatrix}cos(\alpha \plus{} \beta) & \minus{} sin(\alpha \plus{} \beta) \\
sin(\alpha \plus{} \beta) & cos(\alpha \plus{} \beta)\end{bmatrix} \equal{} A_{\alpha \plus{} \beta}$[/hide][/quote]

I see. If I wanted to prove by induction that $ [A_{(\alpha)}]^n \equal{} A_{(n\alpha)}$ a simple induction would suffice, correct?

[quote="triplebig"]
 I see. If I wanted to prove by induction that $ [A_{(\alpha)}]^n \equal{} A_{(n\alpha)}$ a simple induction would suffice, correct?[/quote]
 See $ A_{(\alpha)}\cdot A_{(\beta)}\equal{}A_{(\alpha\plus{}\beta)}$ you can see that with  angles  $ \alpha_1,\alpha_2,...,\alpha_n$
 We have
$ A_{(\alpha_1)}\cdot A_{(\alpha_2)}\cdot .... \cdot A_{\alpha_n}\equal{}A_{(\alpha_1\plus{}\alpha_2\plus{}...\plus{}\alpha_n)}$
It is easy see $ \alpha_1\equal{}\alpha_2\equal{}...\equal{}\alpha_n\equal{}\alpha$ It is $ [A_{(\alpha)}]^n \equal{} A_{(n\alpha)}$

[quote="Kent Merryfield"] Otherwise, it wouldn't be clear what $ \alpha,$ $ \beta,$ or $ \alpha \plus{} \beta$ would mean.[/quote]
 Do you can define $ B_{\alpha} \equal{} \begin{bmatrix}I_s & 0 \\
0 & A_{\alpha}\end{bmatrix} \in M_{s \plus{} 2}(\mathbb{K})$ ? ,$ K \equal{} \mathbb{C}$ or $ \mathbb{R}$ , $ I_s$ is $ s$x$ s$  identity matrix.
 Such that $ B_{\alpha}B_{\beta} \equal{} B_{\alpha \plus{} \beta}$  for angles  $ \alpha$ and $ \beta$
Thanks !


linear algebra
matrix
linear algebra unsolved

Prove that the following three conditions are all equivalent for a real 3x3 symetric
matrix $A$, whose eigenvalues are $a,b,c$

1/ $tr(A)$ is not an eigenvalue of $A$

2/ $(a+b)(b+c)(a+c)\neq 0$

3/ The map $L:S\rightarrow S$ is an isomorphism where $S$ is the set of 3x3 real skew-symmetric

matrices and $L(W)=AW+WA$

1)$\iff$ 2) : is evident: since $tr(A)=a+b+c$ we see that $a+b+c$ not in set ($a,b,c$) $\iff$ $(a+b)(b+c)(a+c)\neq 0$
About 3) i haven't fully understand it yet: you mean that if $L(W)=AW+WA$ the $L$ send $S$ to $S$, it it is, then it also evident $L(W)^{T}=(AW)^T+(WA)^T=-AW-WA$, i think that you meant something else...?

In this case, all that's left to prove is that all antisymmetric matrices (I don't like "skew-symmetric" :)) have the form $AW+WA$. We can use the fact that $A$ is orthogonally diagonalizable, so we can work with $U^tAU$ and $U^tWU$ instead of $A,W$ respectively, where $U$ is an orthogonal matrix which diagonalizes $A$. In other words, we may assume that $A$ is diagonal.

We must thus prove that all antisymmetric matrices can be written as $((\lambda_i+\lambda_j)a_{ij})_{i,j}\ (*)$, where $(a_{ij})_{i,j}$ is antisymmetric. This is clear.

On the other hand, if we asume $3$, we can derive the other parts by using the same idea (if we have $\lambda_i+\lambda_j=0$ for some $i\ne j$, the the antisymmetric matrices we get from must have some $0$ entries which are not on the main diagonal, and this means that we can't get all of them).


inequalities
calculus
derivative
calculus computations

hello ^^
lets $ x$,$ y$y and $ z$ three numbers for $ \mathbb{R}$,find the minimum value for:
$ S\equal{}\Large{\frac{x^2}{x\plus{}yz}\plus{}\frac{y^2}{y\plus{}zx}\plus{}\frac{z^2}{z\plus{}xy}}$

.but what if x,y,z  were getting closer and closer to -1 from below?(infinitessimal)? then wouldnt the expressionn go to zero since the denominator is always larger than the numerator? here is my logic.

for some reason, because of the symetry, i thought that the minimum would be at some point where x=y=z because each can be interchanged with another(not entirely rigorous, please correct if wrong) so i substituted every variable with x, and then simplified to get

$ S\equal{}3(x^2/(x^2\plus{}x))$, which simplifies to
$ S/3\equal{}x/(x\plus{}1)$, and of course minimizing $ S$ is the same thing as minimizing $ S/3$.
so....x/(x+1)=1-1/(x+1). if we want to minimize this, we want to max. the second term, which means we should get the denom as close to zero, meaning x should be as close to -1 as possible...same for y and z b/c of symetry.

[quote]thought that the minimum would be at some point where x=y=z [/quote]

Can you give any mathematical explanation for this ?

The sum can be made arbitrarily large negative- look at $ x\equal{}y\equal{}z\equal{}\minus{}1\plus{}\epsilon$.

This might be more interesting restricted to positive numbers.


Prove that for positive integer $ n,\ 2^{6n \minus{} 5} \plus{} 3^{2n}$ is the multiple of 11.

[quote="kunny"]Prove that for positive integer $ 2^{6n \minus{} 5} \plus{} 3^{2n}$ is the multiple of 11.[/quote]

$ 11|2^{6n \minus{} 5} \plus{} 3^{2n} \Leftrightarrow 11|2^{6n}\plus{}3^{2n}(32)$
$ \Leftrightarrow 11|(\minus{}3)^{2n}\plus{}3^{2n}(32)\equal{}3^{2n}(33)$, which is obviously true.

Well,
[hide]
$ 2^{6n \minus{} 5} \plus{} 3^{2n}$

$ \equiv 2 \cdot 2^{6n \minus{} 6} \plus{} 9^n$

$ \equiv 2 \cdot 64^{n \minus{} 1} \plus{} 9^n$

$ \equiv 2 \cdot ( \minus{} 2)^{n \minus{} 1} \plus{} ( \minus{} 2)^n$

$ \equiv 2 \cdot ( \minus{} 2)^{n \minus{} 1} \plus{} ( \minus{} 2) \cdot ( \minus{} 2)^{n \minus{} 1}$

$ \equiv 0 \pmod{11}$[/hide]

[quote="abcak"]how specific :)

but that should be easy...[/quote]
$ *$.For $ n\equal{}1, 11|11$
So we can assume that for a $ n\equal{}k$ we have:
$ 2^{6k\minus{}2}\plus{}3^{2k}\equal{}11m$

$ **$.For $ n\equal{}k\plus{}1$.We have to prove that:
$ 11|2^{6(k\plus{}1)\minus{}2}\plus{}3^{2(k\plus{}1)}$
But $ 3^{2(k\plus{}1)}\equal{}9 \cdot 3^{2k}\equal{}9(11m\minus{}2^{6k\minus{}2})$

So
$ 2^{6(k\plus{}1)\minus{}1}\plus{}3^{2(k\plus{}1)}\equal{}2^{6(k\plus{}1)\minus{}2}\plus{}9(11m\minus{}2^{6k\minus{}2})\equal{}2^6 \cdot 2^{6k\minus{}2}\plus{}99m\minus{}9 \cdot 2^{6k\minus{}2}\equal{}11(5 \cdot 2^{6k\minus{}2}\plus{}9m)$

[quote="Altheman"](comment: $ 2$ is a generator mod $ 11$)[/quote]
What is a generator (mod 11)?

Some number $ a$ such that $ a, a^2, a^3, \ldots$ forms a complete residue class mod 11.

Is it true that $ a$ is a generator $ \pmod{n}$ if and only if $ \gcd(a,n) \equal{} 1$?

No? Consider a=5, n=6.

$ a^{e}\equiv5\pmod{6}$ for odd e and $ a^e\equiv{1}\pmod{6}$ for even e.

Aren't generators basically equivalent to primitive roots?


What are the different document classes in LaTeX and how are they used?

(e.g. \documentclass{article} can be used to make:

[code]

\documentclass[12pt]{article}
\usepackage{amsmath, amsthm}
\author{name}
\date{\today}
\title{Title}

\begin{document}
\maketitle
\section{Exercise 3-1}
$1+2=3$
\end{document}[/code]

Peter Flynn's [url=ftp://ftp.tex.ac.uk/tex-archive/info/beginlatex/html/chapter3.html#basic]Formatting Information: The Document Class Declaration[/url]

Hi,

have a look at
[list][*] [url=http://www.public.asu.edu/~rjansen/latexdoc/ltx-22.html]Document Class[/url] (LaTeX Hypertext Help)
[*] [url=http://www.math.uiuc.edu/~hildebr/tex/packages.html]LaTeX tips: Document classes and packages[/url]
[*] [url=http://texblog.wordpress.com/2007/07/09/documentclassbook-report-article-or-letter/]\documentclass{book, report, article or letter}[/url] for another short summary.[/list]
The standard document classes are described by most LaTeX books and introductions, you could also just choose an online text from this list: [url=http://texblog.net/latex-link-archive/introduction-guide/]Introductions, Guides[/url].

Stefan


When you look at the results of a poll now, the bars that show (relatively) how many people get proportionately wider as they get longer (the bars undergo a geometric dilation/homothecy rather than merely extend, as they should).

You can view very many samples of this [url=http://www.artofproblemsolving.com/Forum/index.php?f=19]here[/url], unless this is just me.

I'm using Internet Explorer (the current one, I think), and Windows XP.

I'll also include a poll here, so you can look at the bug right in this thread (I've set it to self-destruct in 10 days).  Please vote, so we can get a good specimen here.

EDIT: with one vote, looking at the poll results almost makes my screen look as though I'm using a monitor with 10x15 resolution, or something like that.

EDIT 2: Hmm.  With my first edit, the poll vanished.  I shall recreate it...

In my opinion, the odd thing is the title of the thread is under the poll box. In the old mathlinks, the title is always on the top.

I don't get it. What's the bug?  :maybe:

I don't see anything strange...I'm using Firefox though.

It depends what computer you are on.  The polls look fine for the computer I am on now, but for they look odd (as he says) on my moms older laptop.  It is an IBM thinkpad, I think.

The polls can take up a lot of room on the pages on some computers, and I think this is mainly a problem on older computers, or computers that don't have some plugin (but my mom's laptop can run the classroom).

[quote="dts"][quote="Valentin Vornicu"]I don't get it. What's the bug?  :maybe:[/quote]

Well... it looks like this (to me):

Take a look at one of the answers with no responses.  See how, between the two colored nubs on either end, there's a layer (the part that changes length according to how many votes the answer gets) one pixel wide, and some number of pixels tall?

When I view the polls, that layer doesn't just get wider as the answer gets more responses relative to the other responses, it gets proportionally taller as well.  So, whereas the layer should be six pixels (or however many pixels it is) high, it ends up being six times as high as it is wide; rather than simply extend the layer, it is homothetically dilated.

I'd post a picture, but the only way I'd know how is by copying it into a MS Word document and posting that as an attachment, and I don't really want to do that with the dial-up I'm using.

EDIT: Okay, the layer's height is naturally more like 12 pixels, so when it gets to be, say, two inches long on my screen, it also becomes 2 feet long on my screen.[/quote]
Same here. The results, instead of going horizontally, go vertically and wider, too.

[quote="Valentin Vornicu"][quote="dts"][quote="Valentin Vornicu"]How about posting the picture itself as an attachment?[/quote]

I don't know how.  I'm not exactly an expert at computer matters...[/quote]If you are using Firefox or Opera there's an Add an Attachment link. If you are using IE you should be able to see a Browse button and an Add Attachment button.

Browse to the file, push the add attachment button once you have selected the file and submit the post :)[/quote]

I know how to do [i]that[/i], I just don't know how to take the picture.

[quote="dts"][/quote][quote="Valentin Vornicu"][quote="dts"][quote="Valentin Vornicu"]How about posting the picture itself as an attachment?[/quote]

I don't know how.  I'm not exactly an expert at computer matters...[/quote]If you are using Firefox or Opera there's an Add an Attachment link. If you are using IE you should be able to see a Browse button and an Add Attachment button.

Browse to the file, push the add attachment button once you have selected the file and submit the post :)[/quote][quote="dts"]

I know how to do [i]that[/i], I just don't know how to take the picture.[/quote]
Try pressing the "Prnt Scrn" key on your keyboard. It takes a snapshot of what's currently on your monitor and places it in the clipboard. From there, you can do copy and paste to put the picture in some sort of document.

Press print screen, then open paint application and go to Edit >> Paste. Then save as .jpg and make the dimensions smaller (like 50% for less than 100kB file).

Alright.

If it's okay, though, I think I'll wait until we get high-speed internet (should happen next week).  I'm using 26k dial-up right now, and I really don't want to think about how long that'll take with this internet connection...

[quote="dts"]I'm using 26k dial-up right now, and I really don't want to think about how long that'll take with this internet connection...[/quote]About 8 seconds.

[quote="dts"] When I view the polls, that layer doesn't just get wider as the answer gets more responses relative to the other responses, it gets proportionally taller as well.  So, whereas the layer should be six pixels (or however many pixels it is) high, it ends up being six times as high as it is wide; rather than simply extend the layer, it is homothetically dilated.[/quote]
it looks like that on my computer too.  the bars that show how many people voted for each thing get tall and fat.  it's probably not a huge problem though, if they look fine on other computers.

Yes, this is an IE6 bug.  I will look at it on Monday.

(You do not have to post a screenshot -- I know what's going on.)

[quote="DPatrick"]Should now be fixed.[/quote]

Yep, looks fine now.  Thanks!

Unfortunately, now it looks like each option is doubled...


Gauss
induction
search
geometry
3D geometry
arithmetic sequence
algebra

Find the value of $ 100 \plus{} 100 \plus{} 100 \plus{} 99 \plus{} 99 \plus{} 99 \plus{} 98 \plus{} 98 \plus{} 98 \plus{} \dots \plus{} 3 \plus{} 3 \plus{} 3 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 1 \plus{} 1 \plus{} 1$.

Rofl

How's this factoring?

[hide="So other people can do it and be enlightened"]Basically, you add the numbers 2 at a time, the first one and the last one, so that ALL the sums equal 101, and then find how many sums there are. There are $ \frac {100}{2}\times3$ sums, so the sum is $ 150\times101 \equal{} 15150$.[/hide]

EDIT:Beaten nooo!
EDIT2: YES POINCARE conjecture

i liked textangle's explanation. thank you, textangle.

This reminds me of a method used by the mathematician Carl Friedrich Gauss. When he was a child, his teacher asked him to add all the integers from 1 to 100, and he came up with 5050 because he saw that 100 + 1 = 101, 99 + 2 = 101, etc. and multiplied it by the number of those kinds of pairs.

Oh, yes, I remember reading that. There were 50 pairs, so he got $ 101 \times 50 \equal{} \boxed{5050}$.

You have just to divide dhe sum into three groups that contain the same numbers , since the numbers of each group are in arithmetic progression just use Gauss formula $ {\frac{n(n + 1)}{2}}$ than multiply it by 3.

This problem reminds me of a similar problem:
Find the value of A=100^2 + 99^2 + ... + 2^2 + 1
I need a short solution.
Also try: A=100^3 + 99^3 + ... + 2^3 + 1

[quote="Bachukas"]1. if n=1, then equation is correct ($ 1^2 \equal{} \frac {1*2*3}{6} \equal{} 1$)

If $ 1^2 \plus{} 2^2 \plus{} 3^2 \plus{} ... \plus{} k^2 \equal{} \frac {k(k \plus{} 1)(2k \plus{} 1)}{6}$, then $ 1^2 \plus{} 2^2 \plus{} 3^2 \plus{} ... \plus{} k^2 \plus{} (k \plus{} 1)^2 \equal{} \frac {k(k \plus{} 1)(2k \plus{} 1)}{6} \plus{}$

$ \plus{} (k \plus{} 1)^2 \equal{} \frac {k(k \plus{} 1)(2k \plus{} 1) \plus{} 6(k \plus{} 1)^2}{6} \equal{} \frac {(k \plus{} 1)(k \plus{} 2)(2k \plus{} 3)}{6} \equal{}$

$ \equal{} \frac {(k \plus{} 1)(k \plus{} 2)(2(k \plus{} 1) \plus{} 1)}{6}$ by mathematical induction, equation is proved.

2. if n=1, then equation is correct ($ 1^3 \equal{} (\frac {1*2}{2})^2 \equal{} 1^2 \equal{} 1$)

If $ 1^3 \plus{} 2^3 \plus{} 3^3 \plus{} ... \plus{} k^3 \equal{} (\frac {k(k \plus{} 1)}{2})^2$, then $ 1^3 \plus{} 2^3 \plus{} 3^3 \plus{} ... \plus{} k^3 \plus{} (k \plus{} 1)^3 \equal{} (\frac {k(k \plus{} 1)}{2})^2 \plus{} (k \plus{} 1)^3 \equal{} \frac {k^2(k \plus{} 1)^2}{4} \plus{} (k \plus{} 1)^3 \equal{}$

$ \equal{} \frac {k^2(k \plus{} 1)^2 \plus{} 4(k \plus{} 1)^3}{4} \equal{} \frac {(k \plus{} 1)^2(k \plus{} 2)^2}{4} \equal{} (\frac {(k \plus{} 1)(k \plus{} 2)}{2})^2$ by mathematical induction, equation is proved.[/quote]

That's induction methor. We can use that methor to solve another question.
eg: $ 1\plus{}3\plus{}5\plus{}7\plus{}.............\plus{}(2n\minus{}1)\equal{}n^2$.

[quote="Nguyen Ngoc Linh"]Can you post a pfoof?. I need a proof.[/quote]
Proofs don't belong in classroom math >.>

[quote="Nguyen Ngoc Linh"]Can you post a pfoof?. I need a proof.[/quote]

I believe I derived the formula before. Search the test forums.
Also sum of cubes is $ (\frac{n(n\plus{}1)}{2})^{2}$


Helly's theorem states that given the set of $n$ convex sets s.t. the intersection of any $3$ of these sets is nonempty we can conclude that that in this case the whole collection has a nonempty intersection;
Could you please give a counterexample in case when the sets are not convex for any $n$.

Just pick $n$ copies of the empty set and fill them with elements such that any $3$ copies share one element in common.

thanks,Julien
You example turned out even more evident then i supposed


factorial
function
number theory proposed
number theory

Define in the natural numbers two functions as: $ f(0)\equal{}0.2$, $ g(0)\equal{}0.5$ and $ f(n\plus{}1)\equal{}(n\plus{}1.2)f(n)$ and $ g(n\plus{}1)\equal{}(n\plus{}1.5)f(n)$. Find:

a) The least $ k$ such that $ kf(2008)g(2008)$ is an integer.
b) The largest $ n$ such that $ 3^n|kf(2008)g(2008)$.

I think that $ f(n)$ must have some power of 5 (because some of the terms end in .5) and that $ g(n)$ must have some power of 2 (because some of the terms end in even decimal). So there must be something smaller than $ 10^{2009}$, I think.


$ cosh^{ \minus{} 1} x \equal{} ln(x \plus{} \sqrt {x^2 \minus{} 1})$,
eg.
Express $ cosh^{ \minus{} 1}\frac{5}{3}$ in natural logarithms.

Let $ y \equal{} cosh^{ \minus{} 1}$

$ coshY \equal{} \frac {5}{3}$

$ \frac {e^y \plus{} e^{ \minus{} y}}{2} \equal{} \frac {5}{3}$

$ 3e^{2y} \minus{} 10e^y \plus{} 3 \equal{} 0$

$ (3e^y \minus{} 1)(e^y \minus{} 3) \equal{} 0$

why is $ ln\frac {1}{3}$ ommited as a soln.?

That's because $ \cosh$ is not injective. Try plotting it. You could say the same about $ x^2$ and $ \sqrt{x}$.


limit
real analysis
real analysis unsolved

I could not  understand the factors  :maybe:
can u explain it ?  :(


integration
rotation
calculus
calculus computations

Find the volume of the solid generated when the region enclosed by $y=x^3$, $y=1$, and $x=0$ is revolved about the line $y=1$.

I prefer the use of cylindrical shells.

Move the graph down 1 unit, so
$y=x^3-1$

We therefore need to find the volume of revolution between $1$ and $0$ when rotated around the $x$ axis.

So $V=\pi\int_0^1 (x^3-1)^2dx$
$=\pi[\frac{x^7}{7}-\frac{x^4}{2}+x]_0^1$

So $V=\frac{9\pi}{14} units^3$

EDIT: Sorry, arithmatic error  :oops:


Circles $ C_1$, $ C_2$, and $ C_3$ have radius $ 1$ and centers $ O$, $ P$, and $ Q$, respectively. $ C_1$ and $ C_2$ intersect at $ A$, $ C_2$ and $ C_3$ intersect at $ B$, and $ C_3$ and $ C_1$ intersect at $ C$ in such a way that $ \angle APB \equal{} 60^{\circ}$, $ \angle BQC \equal{} 36^{\circ}$, and $ \angle COA \equal{} 72^{\circ}$. Find $ m\angle ABC$.

Hmm, this problem doesn't make sense... the three circles are mutually tangent to one another externally (since there's only one point of intersection for every two circles), and since all three circles are congruent, shouldn't the three given angles be congruent, too? Maybe I'm misunderstanding the problem...  :maybe:
(a diagram would be helpful)

Perhaps the circles are in space, and not sitting on a plane

The answer i get is $ \frac{\sqrt{5}\plus{}3}{4}$

Its just an application of the cosine law, but here it is anyway
[hide="solution"]
$ \overline{AB}^2 \equal{} 1^2 \plus{} 1^2 \minus{} 2(1)(1)\cos(60) \equal{} 1\qquad(1)$

$ \overline{BC}^2 \equal{} 2 \minus{} 2\cos(36) \qquad(2)$

$ \overline{CA}^2 \equal{} 2 \minus{} 2\cos(72) \qquad(3)$

Now we use these to find $ \angle ABC$

$ 2 \minus{} 2\cos(72) \equal{} 1 \plus{} 2 \minus{} 2\cos(36) \minus{} 2\sqrt {2 \minus{} 2\cos(36)}\cos(\angle ABC)$

$ \therefore 3 \minus{} 2\cos^2(36) \equal{} 3 \minus{} 2\cos(36) \minus{} 2\sqrt {2 \minus{} 2\cos(36)}\cos(\angle ABC)$

$ \therefore \cos(\angle ABC) \equal{} \frac {\minus{}\cos(36)\sqrt {1 \minus{} cos(36)}}{\sqrt {2}} \equal{} \frac {\minus{}1}{4}$

$ \angle ABC \equal{} \cos^{\minus{}1}\left( \frac{\minus{}1}{4}\right)$

It would appear that the answer i gave in my previous post was wrong
[/hide]

EDIT: I just altered this post like 5 times in the last 5 minutes, I'll try clicking the preview button from now on  :rotfl:

[quote="ocha"]$ 2 \minus{} 2\cos(72) \equal{} 1 \plus{} 2 \minus{} 2\cos(36) \minus{} 2\sqrt {2 \minus{} 2\cos(36)}\cos(\angle ABC)$

$ \therefore 3 \minus{} 2\cos^2(36) \equal{} 3 \minus{} 2\cos(36) \minus{} 2\sqrt {2 \minus{} 2\cos(36)}\cos(\angle ABC)$

[/quote]
Shouldn't the LHS be $ 4\minus{}4\cos^2(36)$?

Actually this is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=410870#410870]#7[/url] in The Intermediate Geometry Marathon, but I don't really see how 4everwise's getting 90 degrees as the answer...  :huh:

Could someone please explain why 4everwise's getting $ 90$?  :huh:


ratio
function
geometric series
real analysis
real analysis unsolved

Consider the sum $ \sum_{n\equal{}0}^{\infty}\frac{x}{(1\plus{}|x|)^n}$
Show that the sum converges for all $ x$, but not uniformly.

When $ x\equal{}0$, every term is 0, making the series 0. Otherwise, $ \frac{1}{1\plus{}|x|}$ is strictly between 0 and 1, so the geometric series with it as a ratio converges and we have
\[ x\sum_{n\equal{}0}^\infty \left( \frac{1}{1\plus{}|x|}\right)^n \equal{}\frac{x}{1\minus{}\frac{1}{1\plus{}|x|}}\equal{}\frac{x}{|x|}(1\plus{}|x|)\]
The convergence can't be uniform because a uniformly convergent series of continuous functions is continuous, while this series has a discontinuity (and not even removable!) at 0.


1. Apparently sale of the CD with former physics olympiads has been discontinued. If you still wanted to get those exams, what would you do?

2. What is the next best place or book to find practice physics problems that are similar to the ones on the physics olympiad?

3. Which of the classes listed in this link http://epgy.stanford.edu/courses/physics/ in the third column are most relevant to physics olympiad topics and why?

Dude, there are so many redundant threads about this that it is not even funny.

1. Past exams are impossible to get.  If you want online sources of other good problems, see the thread

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212424

and take feynman's suggestion of BAUPC.  The thread is lower on the page and is quite ambiguously titled "Past Exams."  lol

2. I would recommend Irodov's Problems in Physics.  The thread above also has suggestions.

3. Well, the first two classes in the third column might help, but you might as well just self-study that material if you're just interested in the physics olympiad, and I don't know about EPGY.  But then again, [b]there's more to physics than the physics olympiad[/b]!!

3. Right... there's more to math than math olympiad too but ... I dunno it's like if you were making a choice to learn some number theory or statistics, then number theory would just be more useful in the context of a high school math competition...

i dunno... :|

Also, supposing you would prefer not to self study for various, indisputable reasons that won't be discussed on this thread because that is a digression...

I'm actually a little surprised the the first two would be more relevant than mechanics and electricity/magnetism classes...

1. Why are the first two more relevant?

2.Do the first two classes in the "university" level column contain material you wouldn't see much of in a high school AP physics class?

That's right, what I meant when I said the first two classes are more relevant is that the physics olympiad doesn't require extensive knowledge of advanced topics, like you're not going to need Lagrangian mechanics.  But you will need to know about waves/optics/thermodynamics/some modern physics, which aren't typically covered in AP Physics.  As long as you have a solid understanding of fundamental concepts like those in Halliday and Resnick, and you can apply them creatively, you'll be fine.  Plus, EPGY is kind of expensive.

Taking other classes in the third column isn't necessary for the physics olympiad, but they might be of interest if you want to know more about physics as a whole.  Looking on that website, I find it strange that EPGY requires you to study PDEs concurrently with using Griffith's Introduction to Electrodynamics for their E+M course.


geometry
3D geometry
number theory
greatest common divisor
number theory unsolved

proof that if $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$is an integer so $abc$ is a cube .

It posted before (then $a=b=c\Longrightarrow abc \-cube$).

It's not correct that $a=b=c$.
You see that if $a=1,b=2,c=4$ then $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac12+\frac24+\frac41=5$ is integer.

[i]Proof :[/i]
WLOG suppose that  $gcd(a,b,c)=1$. By problem hypothesis $abc|a^{c}+b^{2}a+c^{2}b$. If $p|a$ then $p|c^{2}b$. So one of $b,c$ are divisible by $p$. WLOG suppose $p|b$ then if $d=gcd(a,b)$ then $d^{2}|c^{2}b\Rightarrow d^{2}|b$. We know that $gcd(\frac{a}{d},\frac{b}{d})=1$. Thus $gcd(\frac{a}{d},d)=1$, also $b|a^{2}c$. Now suppose $\parallel a\parallel _{p}=k,\parallel b\parallel _{p}=s$ then $2\min\{k,s\}\leq s$ and $2k\geq s$. This concludes obviously that $s=2k$. So $\parallel abc\parallel _{p}=3k$. So $abc$ is cube.


Where can I found the solution of All russian mathematics olympiad 2008 on the web?

The web is right here on the [url=http://www.mathlinks.ro/resources.php]ML/AoPS resources section[/url] where you also can find the problems of the [url=http://www.mathlinks.ro/resources.php?c=143&cid=61]5th round of the Russian Math Olympiad.[/url]

[quote="daplikos"]Thanks "orl"
I have a good resource of russian olympiad.But I want to have the solution of them too.specially I want to have the solution of Russian olympiad in 2008 and 2007. I am very happy that you answered my message.[/quote]

In my posting above you can see the problems for 2004 to 2008. Click on the link for the year you like and then on the left you will find the problem numbers of the specific grades. Press on those numbers (i.e. a link) which leads you to the discussion of the problem. Some of the problems may already have solutions, others may not. You need to check it!


Let $x_1\leq{x_2}\dots\leq{x_n}\leq{y_1}\leq{y_2}\dots\leq{y_n}$ be real numbers. Prove that:
\[ (x_1+x_2+\dots+x_n+y_1+y_2+\dots+y_n)^2\geq{4n{(x_1y_1+x_2y_2+\dots+x_ny_n)}  }\]

http://www.mathlinks.ro/Forum/viewtopic.php?t=81252

 :)


I just realized, you can use the Internet Archive to retrieve the Chinese MO problems (and more importantly, the [b]solutions[/b]) off Kalva!   :D

Specifically, see
<http://web.archive.org/web/*/http://www.kalva.demon.co.uk>
(you'll have to copy and paste, since the I can't get the forum to make a link).
Anything on or before 05 April 2004 will do.

Though, the website doesn't cache all the pages during each date. So sometimes you'll need to look through different dates to find the page you want. This makes downloading the entire archive to the harddrive very difficult (unless you have a LOT of time on your hands :rotfl:, in which case we would love it if you could provide the community with a zip file containing all the problems/solutions).

Of course, the same also applies to Tournament of the Towns, and (the cause of all this) Australian Olympiad.

what happened? why cant kalva have the cmo or the tournament of the towns problems?  because i remember when they were there and then suddenly dissappeared. why?

See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=5304

This is too cooool.... thanks for the info!  :P

[quote="ThAzN1"]
This is too cooool.... thanks for the info!  :P[/quote]

It is!  :)

It's really cool to look at the oldes Kalva pages, you can see how it developed etc. In the beginning, there were just a few Newman problems, IMO problems and Putnam problems! So cool.

[quote="billzhao"]This makes downloading the entire archive to the harddrive very difficult (unless you have a LOT of time on your hands :rotfl:, in which case we would love it if you could provide the community with a zip file containing all the problems/solutions).[/quote]

I don't think it is too difficult for anybody experienced with coding. Unfortunately, all I could do were some experiments with WinHTTrack, and they didn't bring much success, but I think it wouldn't be too hard if one knows how to use WGet. Any volunteers?

  Darij


Let $ a,b,c$ be positive numbers satisfying $ abc\equal{}1$. Prove that $ \frac{a\plus{}3}{(a\plus{}1)^2}\plus{}\frac{b\plus{}3}{(b\plus{}1)^2}\plus{}\frac{c\plus{}3}{(c\plus{}1)^2}\ge 3$

[url=http://www.mathlinks.ro/viewtopic.php?p=463670#463670]See here[/url] for this problem and for more links to threads where it has been discussed.


geometry
3D geometry
tetrahedron
trigonometry
AMC
AIME
analytic geometry

Given all the sides of a tetrahedron whats the quickest way to compute the area? (Not a regular tetrahedron.)

Sorry I can't be clearer, but Mildorf told us about a Heron's Formula for tetrahedrons which I promptly forgot. Maybe PM him.

Surface area or volume?

I don't know either  :blush:

Surface area is simple if it's regular: $s^2\sqrt{3}$. If it's irregular, use Heron's, and remember that there are only six distinct edges. All four faces have a combination of those, so there's probably a way to simplify the sum of Heron's in terms of those six edges. If it's volume, it's more difficult, but not too hard. Find the height and the area of the base of that height, which should nearly always be possible, if not always pretty. :P

[quote="JesusFreak197"] Find the height and the area of the base of that height, which should nearly always be possible, if not always pretty. :P[/quote]

The only way I can think of finding the height is Pythagorean Thm. But you only know the hypotenuse(the lateral edge), how can you get the other side?
It's gonna get really ugly with multiple law of cosines or something.

There is a general formula for the volume of a tetrahedron given the sides, but I doubt you will use it for the AIME:

If the edge between the $i$-th and $j$-th vertices is $d_{ij}$ and the volume is $V$, then

$288V^2 = \left | \begin{array} [c]{c c c c c} 0&1&1&1&1\\ \\ 1&0&d_{12}^2&d_{13}^2&d_{14}^2\\ \\ 1&d_{21}^2&0&d_{23}^2&d_{24}^2\\ \\ 1&d_{31}^2&d_{32}^2&0&d_{34}^2\\ \\ 1&d_{41}^2&d_{42}^2&d_{43}^2&0\\ \\ \end{array} \right |$

Courtesy of Mildorf

[quote="mkkool64"]There is a formula given coordinates as well.  If one of the vertices is at the origin and the other three are at the points (x1, y1, z1)=u (x2, y2, z2)=v and (x3, y3, z3)=w then

$V =\frac{1}{6} \left | \begin{array} [c]{c c c} x_{1}&y_{1}&z_{1} \\ \\ x_{2}&y_{2}&z_{2} \\ \\ x_{3}&y_{3}&z_{3}\\ \\ \end{array} \right |$

This was in AOPS volume 2 in the analytic geometry chapter.[/quote]

So basically, what this is called is the "box product" which essentially just means that you're taking the cross product, which is basically taking the area of the base, and then you're dotting it with the third vector to get a volume.

If you didn't divide it by 6 you would get the volume of a parallipiped with three vertices as those three vectors.  However, when you divide by 6 it gives you the volume of the tetrahedron because you have to divide by 2 once because of the fact that it'll be a triangle and 1/3 because its a volume according to the 1/3(area of base)(h) rule, thus multiplying by 1/6.  Its sorta like the 1/2 b*h rule for triangles and rectangles.. or at least thats the way I look at it I guess.


This is exercise 16.5(b) in Munkres if you want to look at his statement of the problem. If T, T' are topologies on X and U, U' are topologies on Y, consider the product topologies V, V' on X x Y induced by T, U and T', U'. If V<V', does that imply T<T' and U<U'? I believe the answer is yes and wrote up a proof to that effect, but I'm not too sure of myself. Am I correct?

Consider that a set $ A\subset X$ is open iff $ A\times Y$ is open in $ X\times Y$.

I am pretty sure that if one product topology is finer than the other, then all that can be proven is that one of the factor topologies is finer than the other product topology's corresponding factor, not necessarily both. That is just my intuition talking, but I'm pretty sure I can type up a proof of that if you want.

Edit, the above post was made while I was typing this post, and it seems to support what I said.

I see where you're coming from, but I can't find an error in my proof. If $ A \in T$, then $ A \times Y \in V$. Since $ V \subset V'$, we have $ A \times Y \in V'$. So $ A \in T'$. This shows $ T \subset T'$, and a nearly identical proof shows $ U \subset U'$. Am I doing something wrong?

True! I was thinking about it wrong, and you are correct. (I was thinking strictly finer instead of just finer.)


The sides $ AB$,$ BC$ and $ CA$ of the triangle $ ABC$ are tangent to the incircle of the triangle $ ABC$ with center $ I$ at the points $ C_1$,$ A_1$ and $ B_1$, respectively.Let $ B_2$ be the midpoint of the side $ AC$.Prove that the lines $ B_1I$, $ A_1C_1$ and $ BB_2$ are concurrent.

let $ K$ be the intesection of $ BB_2$ and $ A_1C_1$ we prove that $ k$ lies on $ (IB_2)$.
since $ AKB$ and $ BKC$ has the same area and $ area(AKB)\equal{}KC_1.AB.sin(<BC_1K)$ and $ area(AKB)\equal{}KA_1.CB.sin(<BA_1K)$ and :
$ sin(<BA_1K)\equal{}sin(<BC_1K)$, we have :$ KC_1.AB\equal{}KA_1.CB$.
now consider the line parallel to $ IA_1$ through $ C_1$, intersect $ IK$ on $ I_1$, so $ \frac{C_1I_1}{IA_1}\equal{}\frac{C_1K}{KA_1}\equal{}\frac{BC}{AB}$ and also $ <I_1C_1I\equal{}<ABC$ so $ ABC$ and $ C_1II_1$ are similar so $ <C_1IK\equal{}<C_1II_1\equal{}<BAC \equal{}180\minus{}<C_1IB_1$.

Dear Mathlinkers,
you can see also
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&t=472317
Sincerely
Jean-Louis


[b]1.[/b] A lot of people ask this question. A good place to start is to try solving the problems in the three high-school forums, and trying some of the problems in the contest section. For problems in the three high-school forums, reading the discussions will be helpful. For problems in the contest sections, after you solve them, read over other people's solutions -- their approaches will be helpful. Don't post a solution there unless it's different from the solutions already presented.

Then there are some helpful books and stuff too... you'll see a lot of names floating around the site.

[b]2.[/b] Being a genius is, in my opinion, not a legitimate goal to have. I think I've already explained to you why I don't think IQ is something to obsess about, and I'll state it again -- what you do with your intelligent is all that matters.

[quote="21 7 15"][i] My questions need the most purest of honesty
1.how can i prepare for my olympiads on this site and the net at large
2.how can i be a genius[/i][/quote]

practice

http://www.artofproblemsolving.com/Forum/resources.php

this is a pretty good place to start


inequalities geometry inradius function trigonometry AMC USA(J)MO

This one is also very nice :

If a, b, c > 0 such that a + b + c = abc, prove that

(1 + a)^(-1/2) + (1 + b)^(-1/2) + (1 + c)^(-1/2) <= 3/2

Indeed, it is nice:

The identity tgA+tgB+tgC=tgA*tgB*tgC is well known for a triangle ABC, so we can replace a, b, c with tgA, tgB, tgC where abc is an acute triangle (acute because the tangents of the angles are positive). 1/(1+(tgx)^2)=(cosx)^2, so our sum in the left part is, in fact, cosA+cosB+cosC=1+r/R<=3/2 because r(the inradius) <= R/2 (R is the circumradius). The last part can also be proved by using the fact that the function cosx is concave on [0, Pi/2].

Indeed. The fact that cos A + cos B + cos C <= 3/2 can also proved by Jensen's inequality as follows. Since the cosine function is concave on [0, Pi/2] we have

cos A + cos B + cos C <= 3 cos ((A + B + C)/3) = 3 cos (Pi/3) = 3/2.

If a condition a + b + c = abc is given then think of tangents !
Similarly, if ab + bc + ca = 1 think of cotangents.
There is a similar expression for cosines, isn't it Grobber :D ?

In fact we have

cos A + cos B + cos C <= 3/2

also for obtuse triangles. To prove this you need a slightly more sophisticated argument. (Beware of nonconvexity/nonconcavity in [0, Pi].)

Wow, nice...geometry certainly makes it all fit nicely. I proved this far less elegantly with the AM-QM inequality and the fact that a+b+c = abc >= V3.

Can you post your solution with AM GM ? Just curious ...  8-)

Ah, never mind...I made a little error in the beginning that conveniently messes up the whole proof...oh I love it when that happens =).

That's what I expected ... 8-)
Maybe this sounds strange but after finding the trig solution (which is obvious if you've seen something like this before (a + b + c = abc)) I tried for a while to prove this with standard inequalities but it didn't work as far as I know.

So if x + y + z + 2xyz = 1 then think of cosines !
(Substitute x = cos A, y = cos B, z = cos C where A + B + C = 180).

In some USAMO - 1999 or 2000 or so - the condition x + y + z + xyz = 4 was given. The you could substitute x = 2 cos A, y = 2 cos B, z = 2 cos C !

erm. could you prove these properties of tangent/cotangent/cosine please? or at least give some hints as to how to prove it...i.e. should we stick with using the geometric defenitions of the trigonometric functions by dropping a couple of perpendiculars, or should we handle it all algebraically with trigonometric identities...or what?

Trig identities ! Just use the following :
sin(180 - A) = sin A
cos(180 - A) = -cos A
tan(180 - A) = -tan A
sin(X + Y) = sin X cos Y + cos X sin Y
cos(X + y) = cos X cos Y - sin X sin Y
tan(X + y) = (tan X + tan Y)/(1 - tan X tan Y)
and some algebra of course ! :D

The following problem is also very nice and can be related to the above posts.

Given 13 distinct reals, prove that there are two of them, a and b, such that
(a - b)/(1 + ab) <= 2 - 3^(1/2).

If you don't know about trig substitution this is too difficult, but now I gave you already the trig hint so it should be easy ! 8-)

Oh ok we use the identities.
OK now, if we let a=tan A and b=tan B, that expression is tan (A-B). Taking our 13 reals mod Pi, some two must be less than or equal to Pi/12 from each other (by pigeonhole). Taking these two to be A and B gives tan (0) :le: tan (A-B) :le: tan (:pi:/12), since tangent is continuous, which gives our desired inequality.
Correct?

That's it !

[quote="MysticTerminator"] Taking these two to be A and B gives tan (0) :le: tan (A-B) :le: tan (:pi:/12), since tangent is continuous, which gives our desired inequality.[/quote]

Do you mean : tan(0) :le: tan(A-B) :le: tan(:pi:/12), since tangent is INCREASING ? ;-)


Prove that: If p is prime than 2^p + 3^p cannot be represented as n^k with natural numbers n and k > 1.

We easily verify that 2 p + 3 p is not of the form a n for some n > 1,  when p = 2,3,5.
Now suppose that p  \geq 7 is a prime. Then  p = 4k+1 or p = 4+3.
Moreover 2 p + 3 p = (2+3) \sum (-1) i 2 i 3 p-1-i , where the sum is over i = 0,...,p-1.
Since 2 = -3mod[5], it is easy to verify that the sum is equal to p mod[5] when p = 4k+1, and to -p mod[5] when p = 4k+3.
Thus, in each case, 2 p + 3 p is divisible by 5 but not by 5 2 . Then, it cannot be a number of the form a n where n>1.

Pierre.

further more about 2^p+3^p: prove that if this number is divisible by some prime q < p then q=5 :)


A square is drawn inside a rectangle. The ratio of the width of the rectangle to a side of the square is $ 2: 1$. The ratio of the rectangle's length to its width is $ 2: 1$. What percent of the rectangle's area is inside the square?

$ \textbf{(A)}\ 12.5 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 87.5$

[hide="Solution"]Let the square's side length be $ s$. Then, the width of the rectangle is $ 2s$, and the length of the rectangle is $ 4s$.

The ratio of the square's area to the rectangle's area is $ \frac {s^2}{(2s)(4s)} \equal{} 0.125 \equal{} 12.5\% \Rightarrow \boxed{A}$[/hide]


Solve for $ x$: $ 3x\minus{}(1\minus{}x)\equal{}5$. Express your answer as a common fraction.

[quote="GameBot"]Solve for $ x$: $ 3x \minus{} (1 \minus{} x) \equal{} 5$. Express your answer as a common fraction.[/quote]

$ 3x \minus{} (1 \minus{} x) \equal{} 5 \implies 3x \minus{} 1 \plus{} x \equal{} 5 \implies 4x \minus{} 1 \equal{} 5 \implies x \equal{} \frac{3}{2}$


that is correct...this is too easy for getting started though

you add 1+3, then 4^2=16, 16-16=0 2*0=0

this should go to middle school, generally, you go parathensis, exponents, mult/div, add/sub

You do the $1+3$ first, then $4^2$, then $16-16$, then $2\times0$

First of all, yo calcule 1+3, then, 4^2, then 16-16 and finally 2*0.

I learned
Please Excuse My Dear Aunt Sally
Each word starts with a letter which also stands for
Parentheses, Exponents, Multiplication, Division, Addition, Subtraction
It is still how I remember order of operations.

How about [hide][b]P[/b]urple [b]E[/b]ggplants [b]M[/b]ake [b]D[/b]elicious [b]A[/b]fternoon [b]S[/b]nacks.[/hide]

 :rotfl:

You start out by following order of operations in each of the expressions. Then once you simplify each thing you get 0=0=0 which is the answer.

I like robinhe's one. I always think of it to help me remember pemdas

I just remeber PEMDAS and think what each thing in PEMDAS means. I don't use any acrostic device.


geometry
geometric transformation
rotation
invariant
symmetry
number theory unsolved
number theory

Suppose that $n\in\mathbb{N}$ is a fix number , Prove that $\forall k\in\mathbb{N}$, \[ \sum_{i=1}^n k^{gcd(i,n)} \equiv 0 (mod n). \]

$\sum_{i=1}^n k^{gcd(i,n)} = \sum_{d|n} \phi(d) k^{n/d}$ is $n$ times the number of neclaces with $n$ beads having $k$ different colors, when two that are identical up to a rotation are considered the same.

why?
$\sum_{i=1}^n k^{gcd(i,n)} =\sum_{d|n} \phi(d) k^{n/d}$
please explain more.

$\gcd(n,i)$ is clearly a divisor of $n$. The question is, how often the exponent $d|n$ occurs in the sum.
Well there are $\phi(n)$ of them for $d=1$, and similar there are $\phi(\frac{n}{d})$ of them in general.
Thus the result follows.

Now the prove that the formula $\frac{1}{n}\sum_{i=1}^n k^{gcd(i,n)} = \frac{1}{n}\sum_{d|n} \phi(d) k^{n/d}$ gives the desired number of necklaces:

Consider any $d$, $1 \leq d \leq n$. There are $k^{\gcd(n,d)}$ necklaces that are invariant under rotation by $d$ (the rotational symmetry by $d$ reduces to the one by $\gcd(n,d)$ by Bezout's theorem and we can choose the first $\gcd(n,d)$ beads as we want).
Now when we add up all those terms, we counted every type (meaning the class of necklaces seen as the same under rotation) necklace exactly $n$ times:
When a type has rotational symmetry of exactly $r|n$ (thus $r$ is the smallest divisor of $n$ such that the necklace is left the same when rotating by $r$), then consider the 'equivalences' given by a rotation by multiples of $r$ only: every such equivalence class is counted $n/r$ times (since this is the number of $d$ with $r|d \leq n$)
There are $r$ truely different classes (those reached by rotation with $d \leq r$) that are considered the same when using all rotations, thus we counted it $n$ times.

Note that this is also a simple corollary to Burnsides (see http://www.mathlinks.ro/Forum/viewtopic.php?t=60832) and Polyas counting lemmas. Here the group of residues $\mod n$ acts cannonical on the mappings $\{1,2,...,n\} \to \{1,2,...,k\}$, and another use of these can be seen in http://www.mathlinks.ro/Forum/viewtopic.php?t=63678 .

In case you want some more general fun:

Let $ K$ be a number field and $ \mathcal O_K$ it's ring of integers. Let $ k \in \mathcal O_K$, let $ \mathfrak n$ be a nonzero ideal of $ \mathcal O_K$ and for an ideal $ \mathfrak m$ of $ \mathcal O_K$ define the norm $ \mathcal N(\mathfrak m) : \equal{} |\mathcal O_K / \mathfrak m|$. Then we have
\[ \sum_{i \in \mathcal O_K/\mathfrak n} k^{\mathcal N((i)\plus{}\mathfrak n)} \equiv 0 \mod \mathfrak n .\]

Note that for $ K\equal{}\mathbb Q$, we get the original one.


inequalities
AMC
USA(J)MO
USAMO
algebra unsolved
algebra

the quesiton:
the sum of 5 real numbers is 8 and the sum of their squares is 16. What is the largest possible value for one of the numbers?

We have $a_1+a_2+a_3+a_4+a_5=8$ and $a_1^2+\ldots =16$ so from that we obtain
$a_1^2+a_2^2+a_3^2+a_4^2+a_5^2-2a_1-2a_2-2a_3-2a_4-2a_5=0$

\Rightarrow $(a_1-1)^2+(a_2-1)^2+(a_3-1)^2+(a_4-1)^2+(a_5-1)^2=5$

So one get's the maximum value if all braces are zeros except one (because all
values in braces are positive so one is maximum when others are all 0)

so $(a_1-1)^2=5$ and from that the maximum value is $\sqrt5 +1$

I hope so that this is correct :) :)

When one actually has a+b+c+d+e = s, a^2+b^2+c^2+d^2+e^2 = f then by applying AMGM you can get
f - a^2 >= 1/4 ( s-a)^2 so that a satisfies the inequality:
5a^2-2as+s^2-4f <= 0 and in this way you can find the boundaries for each of the variables

[quote="alishkov"]When one actually has a+b+c+d+e = s, a^2+b^2+c^2+d^2+e^2 = f then by applying AMGM [/quote]

U can't use AMGM becouse these are real numbers not $positive$ real

you have to get a^2+b^2+c^2+d^2 >= 1/4 (a+b+c+d)^2. Expand it and you get to prove:
3(a^2+b^2+c^2+d^2)>=2(ab+ac+ad+bc+bd+cd). You have that:
3(a^2 + ... ) >= 2( |ab| + |ac| +...) which in turn is grater or equal than 2(ab+ac+ad+bc+bd+cd)
OK ? So in this case AMGM does not need positive numbers

let $a+b+c+d+e=8, a^2+b^2+c^2+d^2+e^2=16\ (a\leq b\leq c\leq d)$
Using Cauchy-Schwalz's inequality,
$(1^2+1^2+1^2+1^2)(a^2+b^2+c^2+d^2)\geq (1\cdot a+1\cdot b+1\cdot c+1\cdot d)^2$

$\Longleftrightarrow 4(16-e^2)\geq (8-e^2)\Longleftrightarrow 0\leq e\leq \frac{16}{5}$

Thus desired answer is $\frac{16}{5}$.

[quote="Tellah"]m@re your solution is wrong because if you calculate you will find that $(\sqrt5+1) +1+1+1+1\not=8$ and neither is the sum of squares 16.(you practicaly find the maximum value of one of the numbers when the sum of the numbers is half the sum of the squares but you forgot that the sums actually have values :8 and 16)[/quote]

I saw that little bit later  :)  :)  :)  :)

This problem was on MIMC contest!!
I just realived I solved a USAMO without realiving it...


One solution of the equation x^3 + 5x^2 + 5x - 2 = 0 is - 2.  Find the sum of the remaining solutions.

Solutions sum to 5, thus the rest sum to 7

and we don't need to know that one is -2

[hide]divide to get x^2+3x-1. The sum of roots is -b/a which equals -3/1= [b]-3[/b][/hide]

[quote="kstan013"][quote="The QuattoMaster 6000"]He meant divide by x+2.  By the way, me@home, the sum of the solutions is -5; not 5.  The sum of the roots of $ax^{3}+bx^{2}+cx+d$ is -b/a.[/quote]No, it's only like that for 2nd degree quadratics[/quote]
That is not true:
[hide] Say the roots of $ax^{3}+bx^{2}+cx+d$ are m, n, and p, so $x^{3}+\frac{bx^{2}}{a}+\frac{cx}{a}+\frac{d}{a}$ also has roots m, n, and p.  This means that $(x-m)(x-n)(x-p)=x^{3}-(m+n+p)x^{2}+(mn+np+mp)x-mnp=x^{3}+\frac{bx^{2}}{a}+\frac{cx}{a}+\frac{d}{a}$.  Corresponding the $x^{2}$s, we get that $\frac{bx^{2}}{a}=-x^{2}(m+n+p)$.  This means that m+n+p=-b/a.  [/hide]

Forget I said anything then. Anyway I suppose it makes sense when you think about the synthetic division process


LaTeX
number theory unsolved
number theory

if [b]p >2[/b] is a prime number and[b] t[/b] is an odd number, [b]4-p <t <p[/b]. prove that there is not exist[b] m[/b] such that :
[b] (p+t)!-t m=p m .   [/b]

If there weren't any mistakes,here's solution.
   With $t$ different from $-p,-p+1,-p+2$ we have $t$ is odd and $m$ is odd as $(p+t)!$ divisible by  $4$
   Hence $(p+t-1)!$=$p^\ {m-1}$ $+...+$ $t^\ {m-1} $,which is odd,contradiction.

Guys, try to learn $\LaTeX$ (it's simple to learn! -read this topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 and test the features in the [url=http://www.mathlinks.ro/Forum/index.php?f=224]Test Forum[/url]), instead of [b]bolding[/b] the equations. :)

Ok [b]Valentine[/b],i've edited my post.

[quote="nalpaction"]Ok $Valentine$,i've edited my post  :D  :D  :D[/quote]It's great you learned $\LaTeX$ so quickly, but please leave text out of the formulae (my name for example). ;)


a,b,c are positive reals show that:

$ \frac{a}{2} \plus{} \frac{b}{3}  \plus{} \frac{c}{6} \geq \sqrt{a}\sqrt[3]{b}\sqrt[6]{c}$

[hide="Hint"]
$ \frac{a}{2}\plus{}\frac{b}{3}\plus{}\frac{c}{6} \equal{} \frac{a\plus{}a\plus{}a\plus{}b\plus{}b\plus{}c}{6}$[/hide]

Kouichi Nakagawa's hint is very good. Now use AM-GM for 6 variables.

[quote="whomer"]Thanks a lot. I'm dumb.[/quote]

If you can't solve one problem, it doesn't mean that you are dumb...

$ a^{1/2}b^{1/3}c^{1/6} \equal{} (a^3b^2c)^{1/6}.$ So now we must prove that
$ \frac{3a \plus{} 2b \plus{} c}{6} \geq (a^3b^2c)^{1/6}.$ Rewriting $ 3a \plus{} 2b \plus{} c$ as $ a\plus{}a\plus{}a\plus{}b\plus{}b\plus{}c$ we see that this is true by $ AM\minus{}GM$ in 6 variables.

Q.E.D.

That solution has already been written. I don't see the purpose of writing it again...


Show that $ \forall x,y,z>0$ we have:
$  \frac{(x+y)(y+z)(z+x)}{4xyz} \geq \frac{x+z}{y+z} + \frac{y+z}{x+z}$.

made by Caragea C.

Here is what I have found:

$ \frac{x+y}{4xyz}= \frac{1}{4yz} +\frac{1}{4xz} \geq \frac{1}{(y+z)^2}+ \frac{1}{(x+z)^2}$ and so we have that:
$ \frac{x+y}{4xyz} \geq \frac{1}{(y+z)^2} + \frac{1}{(x+z)^2}$  and we multiply by $(y+z)(x+z)$ and we have our ineq! :D :D

do you have other solutions? cheers! :D :D

The following inequality is also true.
For all triangle holds: $\frac{R}{r}+1\geq\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$.

[quote="Lagrangia"]Show that $ \forall x,y,z>0$ we have:
$  \frac{(x+y)(y+z)(z+x)}{4xyz} \geq \frac{x+z}{y+z} + \frac{y+z}{x+z}$.

made by Caragea C.[/quote]
refinement 1
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=466241
refinement 2
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=467679

[quote="Lagrangia"]Show that $ \forall x,y,z>0$ we have:
$  \frac{(x+y)(y+z)(z+x)}{4xyz} \geq \frac{x+z}{y+z} + \frac{y+z}{x+z}$.
[hide]made by Caragea C.

Here is what I have found:

$ \frac{x+y}{4xyz}= \frac{1}{4yz} +\frac{1}{4xz} \geq \frac{1}{(y+z)^2}+ \frac{1}{(x+z)^2}$ and so we have that:
$ \frac{x+y}{4xyz} \geq \frac{1}{(y+z)^2} + \frac{1}{(x+z)^2}$  and we multiply by $(y+z)(x+z)$ and we have our ineq! :D :D

do you have other solutions? cheers! :D :D[/hide][/quote]
This inequality is equivalent to
\[sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}(\frac{sinB}{sinC}+\frac{sinC}{sinB})\le\frac{1}{4}\]\[\Leftrightarrow \]\[\frac{R}{r}\ge 2(\frac{b}{c}+\frac{c}{b}).\]

[quote="sandumarin"]Inequality $ \frac{R}{r}\ge\frac{b}{c}+\frac{c}{b} $ belongs V.Bandila (see GM.2/1985,problem C:474 )
With usual notations the following result is true:
1) $ R(\frac{1}{r'}+\frac{1}{R'})\ge\frac{a}{a'}+\frac{b}{b'}+\frac{c}{c'} $
valid for two triangles ABC and A'B'C'
For $ a'=a,b'=c,c'=b $ from 1) to obtain Bandila's inequality.
For $ a'=b,b'=c,c'=a $ from 1) to obtain $ \frac{R}{r}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-1 $
which is stronger than Euler's inequality $ \frac{R}{r}\ge 2 $
_________
Sandu Marin[/quote]

My book P320:
BF151 (d):
\[\frac{R}{r}+\frac{r}{R}-\frac{1}{2}\geq \frac{b}{c}+\frac{c}{b}\]


number theory
greatest common divisor
function
search
group theory
number theory proposed

For each positive integer $ n$ let $ \sigma(n)$ and $ \tau(n)$ be the sum of the positive divisors of $ n$ and the number of positive divisors of $ n$ respectively.

Then \[ \sigma(n)\equal{}\sum_{k\equal{}0}^{n\minus{}1}{\tau(gcd(n,k))}\] holds for any positive integer $ n$.

:)

Using the formula $ \sum_{a|x}\phi(a) = x$,
\begin{eqnarray*}
\sigma(n) & = & \sum_{d|n} d \\
& = & \sum_{d|n} \sum_{e|d} \phi(e) \\
& = & \sum_{e|n} \phi(e)\tau(\frac {n}{e}) \\
& = & \sum_{k = 0}^{n - 1} \tau(\gcd(n,k))
\end{eqnarray*}

Nice proof pluristiq. I know 3 very different solutions including yours, which may be the esiest of them once the initial formula is known. Let's wait some days, if nobody post another solution, I'll post the other two.

It would be iluminating to explain a little more the lats equality, I think.

 :)

Here are the details I left out.

$ \sum_{d|n}\sum_{e|d}\phi(e)\equal{}\sum_{e|n}\phi(e)\tau(\frac{n}{e})$: In the sum on the LHS, $ \phi(e)$ occurs as many times as there are values of $ d$ that satisfy $ e|d|n$, and it's not hard to see there are $ \tau(\frac{n}{e})$ such values.

$ \sum_{e|n}\phi(e)\tau(\frac{n}{e})\equal{}\sum_{k\equal{}0}^{n\minus{}1}\tau(\gcd(n,k))$: Let $ \gcd(n,k)\equal{}d$. Clearly $ d|n$, and for each $ d|n$, the number of times that $ \tau(d)$ occurs in the sum $ \sum_{k\equal{}0}^{n\minus{}1}\tau(\gcd(n,k))$ is the number of $ k$, $ 0\leq k\leq n\minus{}1$ for which $ \gcd(n,k)\equal{}d$, and it's not hard to see that there are $ \phi(\frac{n}{d})$ such values. Thus $ \sum_{k\equal{}0}^{n\minus{}1}\tau(\gcd(n,k))\equal{}\sum_{d|n} \phi(\frac{n}{d})\tau(d)\equal{}\sum_{e|n}\phi(e)\tau(\frac{n}{e})$.

Here is one of the solutions...
[hide]Let $ f$ and $ g$ be functions defined from the positive integers to the complex numbers. For each positive integer $ n$ denote $ \displaystyle\sum_{d|n}{f(d)g\left(\frac {n}{d}\right)}$ with $ (f\times g)(n)$, and with $ f\times g$ the function with those values.

The product just defined is associative and conmutative.

Let $ id$ be the function with values $ 1$ at the number $ 1$ and $ 0$ elsewhere, $ I_k$ be the function with value $ n^k$ at the positive integer $ n$ and  $ \mu$ the Mobius function.

It is easy to see that $ \mu\times I_0 \equal{} id$, $ \tau \equal{} I_0\times I_0$, $ \sigma \equal{} I_0\times I_1$ and $ \phi \equal{} \mu\times I_1$. Also $ id\times f\equal{}f$ for any $ f$.

Taking the product it follows that $ \sigma \equal{} \phi\times \tau$. From here proceed as in plurestiq solution.   [/hide]

:)

The other solution... I think this is very nice, but it isn't completely elementary.
[hide]The idea is a double counting. I guess one could search for a "don't say the word group" equivalent proof but I haven't tooken the time for it.

Consider a cyclic group $ G$ of order $ n>0$ and let $ S$ be the set of all its subgroups. Note that if $ 0<d|n$ there exists a unique $ H$ in $ S$ with $ d$ elements. This implies $ \sigma(n)\equal{}\sum_{H\in S}{|H|}$. The identity follows by counting how many $ H$'s contain a given $ g\in G$. That is how many times a given $ g$ appears in $ \sum_{H\in S}{|H|}$ an adding this numbers over all $ g's$. [/hide]

 :)


ugh. I'm not good at drawing diagrams, so I'll explain the diagram the best I can:

Take a 3*1 rectangle and split it into unit squares, put them together, and shade them. Attach a unit square to the bottom of the leftmost square, and shade that. Then draw nine unit squares adjacent to the sides to the shaded squares, but don't shade them, and then number them 1-9.

The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge to edge. One more congruend square is attached to an edge at one of the nine positions ndicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?

$\left(A\right) 2$
$\left(B\right) 3$
$\left(C\right) 4$
$\left(D\right) 5$
$\left(E\right) 6$


inequalities
function
calculus
derivative
inequalities proposed

If $ a$,$ b$ and $ c$ are positive reals and $ 0\le p \le 1$ be real number prove the following inequality:

$ \frac{a}{(b+c)^{p}}+\frac{b}{(c+a)^{p}}+\frac{c}{(a+b)^{p}}\ge \frac{1}{2^{p}}(a^{1-p}+b^{1-p}+c^{1-p})$

OK, as the inequality is homogenous, we can put a constraint $ a+b+c=1$. Then we can substitute the sums in denominators with $ 1-a, 1-b, 1-c$ respectively, and with the use of Jensen's inequality show that $ LHS{\geq}(3/2)^{p}$. Then you can prove that $ a^{1-p}+b^{1-p}+c^{1-p}{\leq}3^{p}$, again by using Jensen's inequality. In the end we get

$ \frac{a}{(b+c)^{p}}+\frac{b}{(c+a)^{p}}+\frac{c}{(a+b)^{p}}\ge \frac{1}{2^{p}}(a^{1-p}+b^{1-p}+c^{1-p})$

q.e.d.

NOTE: (in the first part of the proof, I've used the fact that the rational functions on LHS are J-convex, and in the second part, I've used the fact that the exponential functions on RHS are J-concave (it can be proven by showing that the sign of the second derivative is positive (and negative, respectively). When we get the last inequality we can forget about the constraint, as the inequality is homogenous again.


algorithm
number theory solved
number theory

Hello, I have a problem in solving the overflow of Modular Exponentiation Algorith. I use VC++ to compute 5^131313 mod 1000000. I defined the numbers as 'unsigned long', which has the range of 0~4294967295. But when the square and multiply algorithm was used, the data wass still too large, therefore it overflowed (I think) and the result was wrong. Does anybody know how to solve it? Thank you!

You have overflow because $(10^6)^2=10^{12}>2^{32}-1\approx 4.29\cdot10^9$.
This leaves you two choices: find a longer format to hold your integers, or factor $10^6$.
The second method: find the number in question mod $2^6$ and mod $5^6$, then solve the chinese remainder problem $x\equiv a\mod 2^6,x\equiv b\mod 5^6$. Since $5^{12}\approx 2.45\cdot 10^7$, these calculations will not cause overflow.

The solution to the chinese remainder problem is $x\equiv 5^6\cdot a\cdot m+2^6\cdot b\cdot n \mod 10^6$, where $m\equiv 5^{-6}\mod 2^6$ and $n\equiv 2^{-6}\mod 5^6$. The calculation $a\cdot m$ can be performed mod $2^6$, and the calculation $b\cdot n$ can be performed mod $5^6$; this avoids potential overflow.

[quote="lynn"] I use VC++ to compute 5^131313 mod 1000000. [/quote]
What a waste of computer power! All you need to do is to note that $5^{131313}\equiv 5\mod 2^6$ (because $5^{16}-1=(5-1)(5+1)(5^2+1)(5^4+1)(5^8+1)$ is divisible by $2^6$ and $131313\equiv 1313\equiv 1280+33\equiv 0+1=1 \mod 16$). So, you need a number divisible by $5^6=15625$ that is $5$ modulo $2^6=64$. But $5^6=125^2\equiv (-3)^2=9\mod 64$, so you need to find a multiple of $9$ that is $5$ modulo $64$. Note that $9\cdot(-7)=-63\equiv 1\mod 64$, so $9\cdot(64-5\cdot 7)=9\cdot 29$ is what you are looking for. The final answer is $15625\cdot 29=15625\cdot 30-15625=468750-15625=453125$. I hope I haven't made some stupid mistake in my computations. My goal was to convince you that, for this particular problem, even using pen and paper is an overkill.

As to the exponentiation algorithm itself, I completely agree with jmerry. As a matter of fact, I once faced a similar overflow problem and had to implement a long integer arithmetic myself. I did it the most stupid way by representing integers as arrays of digits in the system with base $100$. Of course, you'll have to write addition, subtraction, multiplication and division subroutines yourself, which takes time and patience (especially division), but once done, you can use it many times, so it really pays off.

BTW, in your particular situation there is a quick fix. When you multiply two integers $a$ and $b$ both less than $10^6$ and want to get the remainder of the product modulo $10^6$, instead of c=(a*b)%1000000;, write c=(a*(b%1000)+1000*(a%1000)*(b/1000))%1000000; This will keep you within the range up to $2\cdot 10^9$, which shouldn't cause an overflow.

Thank you very much! It's very helpful to me! :)

i implemented some time ago long integer arithmetic in c++.
i used base $2^{16}$ because in this case you can use some asm and make even multiplication and division reasonably fast.
anyway, it took me a long, long time to do it properly  :P


find all fonctions $ f \mathbb{R}\rightarrow \mathbb{R}$ of class $ C^2$ such that :

$ f(x^2)\equal{}f(x\sqrt{2})$

it is a classical fuctional equation!
first for $ y=x\sqrt{2}$ we see that $ f(-y)=f(\frac{(-y)^2}{2})=f(\frac{y^2}{2})=f(y)$ , so $ f$ must be an even fuction.
for $ x > 0$ we have $ f(x^2)=f(x \sqrt{2})$=$ f((x \sqrt{2})^{\frac{1}{2}} \sqrt{2})=....=f(x^{\frac{1}{2^n}}2^{\frac{1}{2^{n+1}}+\frac{1}{2^n}+...\frac{1}{2})}$.
 so for $ n \mapsto +\infty$  since $ f$ is a continious function and $ x^{\frac{1}{2^n}} \mapsto 1$ and $ \frac{1}{2^{n+1}}+\frac{1}{2^n}+...+\frac{1}{2}=1-\frac{1}{2^{n+1}} \mapsto 1$ we get $ f$ is a constant function in positive real set and also by continuity at $ 0$ and $ f$ is even , $ f=c$ for some real number $ c$ which it is clear solution to this problem.


How do you find the continued fraction expansion of a radical. For example $ \sqrt{7}$?

There is a very straightforward [url=http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Continued_fraction_expansion]algorithm[/url] (a simplification of the general continued fraction algorithm) in the case of quadratic surds.  Do you understand how to compute the continued fraction expansion in general?


function
limit
real analysis
real analysis unsolved

Let $ V_{a}^{b}f$ be the [url=http://en.wikipedia.org/wiki/Bounded_variation#BV_functions_of_one_variable]bounded variation[/url] of a function.

Now suppose $ f_{n}\rightarrow f$ uniformly on an interval $ [a,b]$.  Is it necessarily true that $ \lim_{n\rightarrow\infty}V_{a}^{b}f_{n}\equal{}V_{a}^{b}f$?

In our analysis section, we tried to come up with some counterexample and we came up with the function $ f_{n}(t)\equal{}\frac{1}{n}\sin(nt)$, which uniformly approximates the function $ f(t)\equal{}0$.  However, there was a disagreement as to whether or not $ V_{a}^{b}f_{n}\rightarrow 0$.  So does it?

this is of course false. You can find a lot of continuous functions $ f$ such that $ \|f\|_{\infty}<\epsilon$ and the variation of $ f$ over $ [a;b]$ is as large as you want (or even infinite).

We assumed it was false.  We had some disagreements as to whether or not my function worked.

It does: you have about $ \frac{b\minus{}a}\pi n$ humps of height $ 1/n$, so the total variation is at least some fixed constant, which is enough to conclude that it does not tend to $ 0$.

And if we make the example $ f_n(x)\equal{}\frac1n\sin(n^2x),$ then we can have $ f_n$ tending uniformly to zero while $ V_a^bf_n\to\infty.$

(Just making concrete something alekk already said above.)


probability
articles
conditional probability

Your friend has two children. You meet one of them and find out that he is a boy. What is the probability that the other child is a girl?

P.S. - A child can only be a boy or girl

[hide="Solution"]Assume WLOG that the probability of having a boy is equal to the probability of having a girl. (Not explicitly stated; you wanted an elaborate solution.)

Therefore are three cases:

Case 1: The family had a boy first, then a girl.

Case 2: The family had a girl first, then a boy.

Case 3: The family had two boys.

Now, each case has the same probability of arising. Therefore, two cases out of three state that the other child is a girl, and one case out of three states that the other child is a boy. Therefore, the probability that the family has a girl is $ \displaystyle\frac {2}{3}$.[/hide]

See http://en.wikipedia.org/wiki/Conditional_probability  and http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

[hide="Different Solution"]As mentioned in the above wikipedia article, the probability is actually $ \frac{1}{2}$! However, if your friend told you that one of his children was a boy, then the probability would be $ \frac{1}{3}$. This is because finding that the child you met is a boy does not affect the probability that the child you didn't meet is a girl.

If the probability was not $ \frac{1}{2}$, then if the child you met said she was a girl, the probability that the other child is a girl would be different than if the child you met said he was a boy. However, it is intuitive that this is not the case.

You can also think of it this way: there are four possibilities for your friend's children, where the first child is the one you met:
BB BG GB GG
Since the child you met said he was a boy, this rules out the last two possibilities, and thus the probability that the other child is a girl is $ \frac{1}{2}$. However, if your friend told you that one of his children was a boy, only the last possibility would be ruled out, and the probability that the other child is a girl would be $ \frac{2}{3}$.

Finally, as Walk Around The River suggested, we can use conditional probability to get the answer. Assume you met one of the children. Let $ A$ be the event that the child you met is a boy, and $ B$ be the event that the child you didn't meet is a girl. By the rules of conditional probability:
$ P(B|A)\equal{}\frac{P(B\&A)}{P(A)}$
We know that $ P(A)\equal{}\frac{1}{2}$. Furthermore, since $ P(B)\equal{}\frac{1}{2}$ and the events $ A$ and $ B$ are not correlated, $ P(B\&A)\equal{}\frac{1}{4}$. Thus
$ P(B|A)\equal{}\frac{1/4}{1/2}\equal{}\frac{1}{2}$

This is a very tricky question, and I may not have explained it very clearly. Feel free to ask if you have any more questions![/hide]


Does anyone have the answers to the 2005 AMC 10 B?

I did the problems in the wiki but I didn't realize that there were no solutions posted.
I just want to see how much I would have scored, so I don't need the solutions.

Thanks in advance.

See the [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43]Resources Section.[/url]

See also:

http://www.unl.edu/amc/e-exams/e5-amc10/e5-1-10archive/2005-10a/05itemdiff10.html

Steve Dunbar
MAA Director, American Mathematics Competitions


[quote="cincodemayo5590"]Five circles $\omega_{1}, \omega_{2}, \ldots \omega_{5}$ are drawn so that the radius of each circle is equal to $16,8,4,2, \text{ and }1$ respectively; the centers the of the circles are collinear in the order given; and the center of each circle, other than $\omega_{1}$, is on the circle which has the closest, but larger, radius than the circle to be drawn ($\omega_{2}$ is on $\omega_{1}$, etc.) Find the common area of the five circles.[/quote]

[hide]Region between $\omega_{1}$ and $\omega_{2}$ has the area of
$4 \cdot \frac16 \cdot 64 \pi-2 \cdot 64 \cdot \frac{\sqrt{3}}{4}$ $=\frac{128}{3}\pi-32 \sqrt{3}$.

Because the pattern repeats, we can add the area of the circles, and subtract the intersections repeatedly.

First the area of the circles sum up to
$16^{2}\pi+8^{2}\pi+4^{2}\pi+2^{2}\pi+1^{2}\pi=341 \pi$.

The intersections has the pattern that continues while the area decreases to $\frac{1}{4}$. So
$(1+\frac14+\frac{1}{16}+\frac{1}{64})(\frac{128}{3}\pi-32 \sqrt{3})$ $=\frac{170}{3}\pi-\frac{85}{2}\sqrt{3}$.

Subtract the intersections from the area of the circles:
$\frac{853}{3}\pi+\frac{85}{2}\sqrt{3}$.[/hide]

That doesn't seem very right though :blush:


1. A book weighing 3N is resting on a table 1 meter above the floor. How much force is the table exerting on the book?

2. How much work is the table doing on the book?

1. If the book is in equilibrium, then the sum of the forces that act on it must be zero. The only forces applied to the book are its weight and the force applied by the table, which then must have the same intensity of the weight (3 N), same direction and opposite sense.

2. Work is a way of tranferring energy by means of some kind of movement. The force that the table exerts on the book is not making it moving, and so the work is zero.