Find the hidden emoticon!
[hide="."]
[hide="."]
Nope!
[/hide]

[hide="."]
Not here either
[/hide]

[hide="."]
[hide="."]
Wrong!
[/hide]
[hide="."]

 :ninja:

[/hide]
[/hide]
[hide="."]

[/hide]

[hide="."]
Time remaining: $ \frac{42}{0}$
[/hide]
[hide="."]
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It would be much harder if there were one on each line...

lol?

message is too short

btw i did find it

[quote="zephyredx"] :ninja: [/quote]

I see.

Wondering if this could start a new kind of forum game...although it's pretty easy to cheat.

[quote="zephyredx"]Find the hidden emoticon!
[hide="a"]
[hide="b"]
Nope!
[/hide]

[hide="c"]
Not here either
[/hide]

[hide="d"]
[hide="e"]
Wrong!
[/hide]
[hide="f"]

 :ninja:

[/hide]
[/hide]
[hide="i"]

[/hide]

[hide="j"]
Time remaining: $ \frac {42}{0}$
[/hide]
[hide="k"]
[/hide]
[/hide]

It would be much harder if there were one on each line...[/quote]

Easy way to cheat. Quote and change.

I'm making a mega-maze, complete with three levels.  The object is to find the message.  First one to post the message wins.

Update: Two Levels are Complete.

It's ready, but somewhat glitchy.

Remember, the first one to post the message hidden inside Level Three wins.

[hide="Start Here!"][hide="Continue Here!"][hide="Hmmm..."]You Lose![/hide]
[hide="Hmmm..."]You Lose![/hide]
[hide="Hmmm..."]You Lose![/hide]
[hide="Hmmm..."]You Lose![/hide][/hide]
[hide="Continue Here!"]You Lose![/hide]
[hide="Continue Here!"][hide="Hmmm..."]You Lose![/hide]
[hide="Hmmm..."]You Lose![/hide]
[hide="Hmmm..."]You Lose![/hide]
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[hide="Continue Here!"][hide="Hmmm..."]You Lose![/hide]
[hide="Hmmm..."]You Lose![/hide]
[hide="Hmmm..."]You Lose![/hide]
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[hide="Continue Here!"][hide="Hmmm..."]You Lose![/hide]
[hide="Hmmm..."]You Lose![/hide]
[hide="Hmmm..."]You Lose![/hide]
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[hide="Start Here!"][hide="Continue Here!"][hide="Hmmm..."]You Lose![/hide]
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[hide="Continue Here!"][hide="Hmmm..."]You Lose![/hide]
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[hide="Hmmm..."]You Lose![/hide]
[hide="Hmmm..."]You Lose![/hide][/hide]
[hide="Continue Here!"][hide="Hmmm..."] :ninja: Congratulations.  You have found the first ninja. You Lose!
You Lose!
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You Lose!
Congratulations.  You have found the second :ninja:. [url=http://artofproblemsolving.com]Mwahahahahaha![/url]
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pwning chickens of doom
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You Lose!
You Lose!
You Lose!
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You Lose!
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You Lose![/hide][/hide]
[hide="Hmmm..."]You Lose![/hide]
[hide="Hmmm..."]You Lose![/hide]
[hide="Hmmm..."]You Lose![/hide][/hide]
[hide="Continue Here!"]You Lose![/hide][/hide]
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[hide="Start Here!"][hide="Continue Here!"][hide="Hmmm..."]You Lose![/hide]
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[hide="Continue Here!"][hide="Hmmm..."]You Lose![/hide]
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[hide="Continue Here!"][hide="Hmmm..."]You Lose![/hide]
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the phrase is pwning chickens of doom

i can't believe i actually went through all that.....
and i can't believe u actually wrote all that....

The answer: pwning chickens of doom

For the last level, it was the last tag I looked. I looked in random order.

good job with the maze.  it was pretty ez, but it was cool.  i like ninjas too.

To ernie: can I do a sequel? It will have a branching story.

[quote="dragon96"][quote="james4l"]pwning chickens of doom[/quote]

<sarcastic><emotionless>mwahahahahaha</emotionless></sarcastic> I beat you to it!!![/quote]

simulpost

[quote="dragon96"][quote="ernie"]I am probably [b]not[/b] making a sequel.  Sorry, james.[/quote]

Oh by the way, his name isn't James. In fact, it isn't even close to James. In fact, it doesn't even start with a J.[/quote]

Yeah, but I don't know his real name.  And I don't want "James" to think that I'm a stalker.

My real name isn't ernie.  :P

[quote="james4l"][quote="james4l"]I said that I wanted to make the sequel[/quote]

well can I make the sequel???[/quote]

If your heart desires it.

Ok everyone I'll probably be done with it in... about 3-4 days...

If you really want to make a maze, BBCode might not be the best idea... but it's the only thing that works on the forum...

[size=9][hide="."]You win![/hide][/size]

dunno about anyone else, but i was WAY too lazy to finish that maze. must hav taken quite a while...

[quote="zephyredx"]Find the hidden emoticon!
[hide="."]
[hide="."]
Nope!
[/hide]

[hide="."]
Not here either
[/hide]

[hide="."]
[hide="."]
Wrong!
[/hide]
[hide="."]

 :ninja:

[/hide]
[/hide]
[hide="."]

[/hide]

[hide="."]
Time remaining: $ \frac {42}{0}$
[/hide]
[hide="."]
[/hide]
[/hide]

It would be much harder if there were one on each line...[/quote]
Easy. Use quote to cheat.

This is level 1

[hide="Level 1"]You are in a forest. What do you do?
[hide="Go north"]
You sense a feeling of victory.
[hide="Go north"]
The sense of victory is getting stronger.

The sense of victory is diminishing.

A portal has warped your location!

A Dire Wolf is attacking!

Your HP = 3/20

You died... sorry

You escaped, you see a walled fortress

Hopefully you knock on the door.

Nah, noone's inside. You need a lockpick to get in.

You found a lockpick! But then the dire wolf attacks you, you die...

HP = 9/20

HP = 15/20

HP= 20/20, max health

You encounter dire wolf, lv 3. You died...

A wall blocks your path.

It goes without saying but you are trapped.

Yes, you go north. But what now?

Ok. The temperature is dropping.

A wall blocks your path.

Where in the world are you? AHHHHHHHH IM FALLING THROUGH A PORTAL!!!!!!!!!!!!!

You are out of the portal. Strangely you are shivering, almost to death.

What? How in the world? You can't go north.

You see nothing interesting...

Good. You keep going north...

A polar bear attacks! You die...

A polar bear attacks! You die...

You are still freezing. You are paralyzed.

You know you're still paralyzed?

You know you're still paralyzed?

A polar bear attacks! You die...

Nothing interesting. But you can't seem to go that direction.

??? attacks! You die...

A rope ladder unravels!

keep on going: elevation = 30 ft

keep on going: elevation = 60 ft

keep on going: elevation = 90 ft

keep on going: elevation = 130 ft

keep on going: elevation = 160 ft

keep on going: elevation = 200 ft

keep on going: elevation = 250 ft

keep on going: elevation = 330 ft

keep on going: elevation = 400 ft

keep on going: elevation = 570 ft

keep on going: elevation = 690 ft
Is this ladder ever going to end?

keep on going: elevation = 820 ft

keep on going: elevation = 1000 ft
I guess not.

keep on going: elevation = 1200 ft

keep on going: elevation = 1500 ft

keep on going: elevation = 1900 ft
I'm getting a bit tired.

keep on going: elevation = 2200 ft

keep on going: elevation = 2800 ft

keep on going: elevation = 4000 ft

keep on going: elevation = 6000 ft
The air is getting a bit thin.

keep on going: elevation = 9000 ft
You died of suffocation....

Yay. You won. To get to the next level: Purchase the "Deluxe Sword"

You trip, fall out of a cliff. You die.

[/hide]
[hide="Go south"]
Suddenly, a rock blocks your path!
[/hide]
[hide="Go east"]
Is it just me or are you going into danger?

Is it just me or are you going into danger?

Is it just me or are you going into danger?

Is it just me or are you going into danger?

Ok, if you insist.

You landed into a bed of cotton. Wow. Yay for luck.

Yes, cotton.

Is there anything I need to say more?

COTTON! COTTON, I SAY! COTTON!

JUST MILES AND MILES OF COTTON!

EVERYTHING IS COTTON! YOU SEE AT LEAST A CUBIC MILE OF COTTON!

FOR ACRES AND ACRES AND ACRES YOU SEE NOTHING BUT COTTONBALLS!

I think this is getting a bit futile.

Maybe...

but you seem to persist.

Didn't I tell you? There's so much cotton...

That you died from suffocation...

[/hide]
[/hide]
[hide="Go west"]
You see a bear HP = 20/20, ATK = 6
[hide="Attack"]
HP_bear = 11/20
HP_you = 14/20
[/hide]
[hide="Run Away"]
Your HP is now 14/20.

You killed the bear! But then its family devours you. You died...

[/hide]
[/hide]
[hide="Go east"]You are blocked by trees.[/hide]
[/hide]

[quote]Yay, you won. To go to the next level purchase the "Deluxe Sword"[/quote]

So, I purchase the Deluxe Sword. Where's the second level?[/quote]

A minor note:
[quote="james4l"]keep on going: elevation = 9000 ft
You died of suffocation.... [/quote]
I've been higher than 9000 feet (to about 13,000 feet in Colorado) and not died of suffocation.  If you meant 9000 meters, now that's a different story.
(9000 meters is roughly equivalent to 30,000 feet, which is taller than Mt. Everest.)

Looking forward to Level 2.

oh sorry yeah there are a few bugs. == should have said died of exhaustation

um... north should have said "you have been pwned by a grue"


induction
inequalities proposed
inequalities

$a,b,c\in \mathbb{R},a+b+c=abc=a^3$.Prove the following:
a)$|a| \geq \sqrt3$
b)$a^{2n} +b^{2n} +c^{2n} \geq 3^{n+1},n \in \mathbb{N}$
The first point is easy..and for the second point I used induction and it came out nicely..but I know there`s gotta be a simpler solution..

the first is easy using $(b + c)^2 - 4bc \geq 0$
 i don't think there is any other way except induction for the second.
 it looks pretty straightforward, so
 i don't see why you want a shorter method  ;)

Well because for the induction step it came out with a discution of the parity of $n$ and there was a lot of writing..Now I wonder if I remember the problem corectly..

:oops: I`ve edited it and now it`s corect..

the second is easy .

b)$a^{2n} +b^{2n} +c^{2n} \geq 3\sqrt[3]{(abc)^{2n}}=3\sqrt[3]{(a^{2n})^3}\geq 3^{n+1}$

with the help from a)


Treething is 69 inches tall. If Midget Joe, who is three times as short as Albatross, standing on Treething's eye is 6.9 feet high and Albatross standing on Treething's eye is 9.6 feet high, what is the distance between Treething's eye and the bottom of his foot in inches?

[hide]Draw a picture. Treething is 5.75 feet high, and Midget Joe goes up to 6.9 feet. Albatross goes up to 9.6 feet. The distance between the top of Treething and the top of JOe is 1.15 feet, and the distance between the top of Treething to the top of Albatross is 3.85 feet. Call the distance from Treething's eye (uncomfortable) to the top of Treething $n$. Then $3(1.15+n)=3.85+n$. Solving for $n$, we get $n = 0.2$ feet. Thus the distance from the ground to Treething's eye is $5.75-0.2 = 5.55$ feet = $66.6$ inches?[/hide]

[hide="..."]Let the distance from Treething's eye to the top of his head be x. Midget Joe's height is 6.9-5.75+x, and Albatross' height is 9.6-5.75+x. Albatross' height is three times Midget Joe's height, so 3(1.15+x)=3.85+x. We solve for x and find that it is .2 ft. Subtracting that from Treething's height, we get the distance from his eye to his feet. The distance is 5.75-.2=5.55 ft, or 66.6 inches.[/hide]

[quote="anirudh"][hide="..."]66 inches
 :rotfl: [/hide][/quote]

it's much better if you give the exact answer like jclarke did


HMMT Putnam limit induction inequalities

What is the value of $ \sqrt {1 \plus{} 2\sqrt {1 \plus{} 3\sqrt {1 \plus{} 4\sqrt {1 \plus{} 5\sqrt {1 \plus{} \dots}}}}}?$

Find some way to solve:
 $ \displaystyle{x \equal{} \frac {\left(\frac {\left(\frac {x^2 \minus{} 1}{2}\right)^2 \minus{} 1}{3}\right)^2 \minus{} 1}{4\cdots}}$

The answer is $ 3$. See [url]http://mathworld.wolfram.com/NestedRadical.html[/url], eq. 26 to 28.

Taking a break from coding...

This problem also appeared as A6 from Putnam 1966. The solution provided reads:

We understand the statement to mean that \[ 3=\lim_{n\rightarrow\infty}\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots\sqrt{1+(n-1)\sqrt{1+n}}}}}.\]We see that \begin{eqnarray*}3&=&\sqrt{1+2\cdot4}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+3\sqrt{25}}}\\ &=&\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{36}}}}.\end{eqnarray*}This leads us to conjecture the relation \[ 3=\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots+\sqrt{1+n\sqrt{(n+2)^2}}}}}\]for all $ n\ge 1$. Proceeding by induction we verify that \[ (n+2)^2=n^2+4n+4=1+(n+1)(n+3)=1+(n+1)\sqrt{(n+3)^2}.\]This given, we must have \[ 3\ge\sqrt{1+2\sqrt{1+3\sqrt{\cdots\sqrt{1+(n-1)\sqrt{(1+n)}}}}}.\]To set an inequality in the other direction, observe that for any $ \alpha>1$\[ \sqrt{1+n\alpha}\le\sqrt{\alpha}\sqrt{1+n}.\] A repetition of this inequality gives then \[ 3\le(n+2)^\alpha\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots\sqrt{1+(n-1)\sqrt{1+n}}}}},\] where $ \alpha=2^{1-n}$.

Kind of a nooby question but is there a way to do it without assuming the answer is 3 and then proving it?


TI-89 Titanium. The best graphing calculator by TI. :lol:

Just a question:
Wheres the 73?

I use a 89 and 73

Or the 82. I used it till about 2 weeks ago, and it goes faster since I don't have so many prgms on it.

I use the 84+Silver now.

Billy

I use my rusty, excuse me , [i]Trusty[/i]
TI-83+ Silver.
I need to get an 89 pretty soon.

i'm a freshman now, but when i competed in MC i always used a TI 89 and my good old TI 34II. those ones are awesome!

TI 34II? Whats that?
*looks it up at TI home*
*gasps in the fact that it doesn't graph*
*shakes his head for ever thinking that all calcs graphed and that you needed a graphing one to get even a 2 in target round*
*laughs at himself*

*hugs the TI-83 Plus I got a Nats last year*

TI-84 silver. Don't we get a TI-89 titanium for competing in Nationals???

Yeah, I've been hearing about that all over the place.  I think we do get a TI-89 Titanium for going to nationals.

YAY. Why don't they give them out before the competition?

Really? A TI-89? Sweet.

biffanddoc, I think it's to prevent us from doing well :P .

I heard you get an 83 for being in the nationals, but I might be wrong.

HP 48 GX, a calculator without equal.

I think Texas Instruments gives SOME calculator to the students at Nats each year, as part of its role as a national sponsor of MATHCOUNTS, but not always the same kind of TI calculator.

[quote="Farazy"]Hey...I have a TI-34II... :?[/quote]

awesome. i used my 34II a lot more for the simple computations and such as i was more familiar with it and it's functions. it's the calculator that i mainly rely on anyways even now. i try to not used my graphing so much so i can do the things on my own...unless i get too lazy. :)

I'm reasonably sure that MATHCOUNTS Nationals gives out TI-83 (either Plus or Silver, I forget) one year and then TI-89 (I think we get Titanium this year though :) good thing I made it :)) the next and then TI-83... (repeats itself)

TI 83 plus--Hopefully I get a Titanium at nationals *fingers crossed*

Ok, I have had kids go to Nats the last 7 years, officially here are the calculators awarded for the last 7 years.

2004 TI-83+ SE
2003 TI-89
2002 TI-83+ SE
2001 TI-89
2000 TI-89
1999 TI-89
1998 TI-92

Now that I think about it the 1998 & 1999 might be reversed.  I had a kid who make it both years, and he got one of each, although I can't remember which came first.

I was probably the only person EVER to have gone to Nationals without a graphing calculator. (I'd never used one before. It was embarrassing to see everybody else with graphing calculators... *cough*) But then again, I probably ranked first in ascending order of Mathcounts preparation hours in the history of the Nationals...

Back on topic, I'm currently smitten in love with my 89 from '03.

I have a TI - 30XIIS.  its scientific, not graphing

[quote="ln(dx/dy)"]Who said "None of the Above"?[/quote]

I don't know, but tinkertoy calculators are cool to use.

I used to do that...  When I went to math competitions in our county during sixth and seventh grade I went with a dinky calculator or none at all... It was more fun and a bit more of a challenge.

 :D

I use the HP32IIS and sometimes the HP33. I always make sure I have a graphing calculator in the target round too. I prefer the simple TI83-Plus.

Haha Dennis, It's the HP32SII and I absolutely HATE the HP33S. The HP32SII is where it's at... and I almost never use graphing calculators...

Yea, I hate the HP33S too so I usually use the HP32. When you use the HP33S, are some of the buttons insensitive? As in, when you press a button (especially the operation signs), does the calculator not register it and nothing happens?

Graphing calculators are nice since you can find the intercepts of lines and roots of polynomial equations, but that doesn't show up on Mathcounts very much.

Yeah the HP made the HP33S have hard-to-press buttons like you have to press harder so when you're putting in something really fast sometimes it will not register a button.

I have a TI-83 Plus.  I wish I had a TI-89 because it can differenciate equations.

I don't care what people say, Texas Instruments makes better calculators than HP.  I only like HP computers and printers.

[hide]yeah I have a TI-34II but I  started mathcounts less than two years ago[/hide]

i also use a ti-83 plus silver, but i tend to steal my brothers ti-89 a lot. but for my birthday i want a ti-89 titanium silver addition. *dream* *sigh*

I don't believe that there is a TI 89 titanium silver edition.  It's only TI 89 titanium.


geometry
3D geometry
trigonometry
function

Thanks. These are the ones I still do not know how to do.

3)Quadrilateral ABCD has right angles at B and D. If ABCD is kite shaped with AB-AD=20 and BC-CD=15, find the length of a radius of the circle inscribed in ABCD.

5) Let S be the infinite sum of $ \dfrac{1}{2} \plus{} \dfrac{1}{2} \plus{} \dfrac{3}{8} \plus{} \dfrac{1}{4} \plus{} \dfrac{5}{32} \plus{} \dfrac{1}{8}...$
Let K be the sum of the squares of the roots of $ x^5 \minus{} 3x^4 \plus{} 6x^3 \minus{} 3x^2 \plus{} 1 \equal{} 0$
Find the product of S and K.

8) ABE is a right triangle with right angle E. C is a point on AE. D is a point on AB. CD forms a perpendicular line to AE. $ AC \equal{} 8, AD \minus{} 10, CD \minus{} 6,$ and  $ < BCD \equal{} 30 degrees.$ If BD= $ k(4\sqrt {3} \plus{} 3)$, find k.

11) Find the area of the triangle with vertices: (1,-3,-2) (5, -1, 2) (-1,1,2)

12) Given $ f(x) \equal{} x^5 \minus{} 5x^4 \plus{} 13x^3 \plus{} 121x^2 \minus{} 25x \plus{} 69$, find the sum of the cubes of its roots.

13) Two poles are 50 feet apart on level ground. One pole is 10 feet taller than the other. A cable hangs form the top of the poles in the shape of a parabola. It's lowest point is 10 feet above a point on the ground 20 feet from the shorter pole. Find the height of the taller pole.

14) The phase shift of $ y \equal{} 5\sin3x \minus{} 12\cos3x$ is:

19) Find the remainder when $ 19^{2001}$ is divided by 23.

As soon as you get one, go ahead and post the solution.

[quote="brianb"]Hey yall here are some problems to solve:

Please explain and work fully:
(These are to be worked with no calculator, and in a method that can be feasibly used in a short amount of time, at least 2-3 minutes, preferably in 60 seconds or less)
Most will be somewhat formula based.

1) Find $ (.99)^{10}$ to the first three decimal places.

2)Find $ x^2 \plus{} y^2 \plus{} z^2 \equal{} a^2$ and $ x \plus{} y \plus{} z \equal{} x^3 \plus{} y^3 \plus{} z^3 \equal{} a$, write an equation expressing $ xyz$ in terms of $ a$.

3)Quadrilateral ABCD has right angles at B and D. If ABCD is kite shaped with AB-AD=20 and BC-CD=15, find the length of a radius of the circle inscribed in ABCD.

4) What kind of graph is $ r \equal{} 1 \plus{} 2\sin\theta$

5) Let S be the infinite sum of $ \dfrac{1}{2} \plus{} \dfrac{1}{2} \plus{} \dfrac{3}{8} \plus{} \dfrac{1}{4} \plus{} \dfrac{5}{32} \plus{} \dfrac{1}{8}...$
Let K be the sum of the squares of the roots of $ x^5 \minus{} 3x^4 \plus{} 6x^3 \minus{} 3x^2 \plus{} 1 \equal{} 0$
Find the product of S and K.

6) Find the remainder when $ 1! \plus{} 3! \plus{} 5!... \plus{} 49!$ is divided by $ 18$.

7)What is the $ 7th$ pentagonal number?

8) ABE is a right triangle with right angle E. C is a point on AE. D is a point on AB. CD forms a perpendicular line to AE. $ AC \equal{} 8, AD \minus{} 10, CD \minus{} 6,$ and  $ < BCD \equal{} 30 degrees.$ If BD= $ k(4\sqrt {3} \plus{} 3)$, find k.

9) The graph of the polar equation $ \cos\theta \plus{} r \equal{} \frac {1}{r}$ is a:

10)Evaluate: $ 2\arctan\frac {1}{3} \plus{} \arctan\frac {1}{7}$

11) Find the area of the triangle with vertices: (1,-3,-2) (5, -1, 2) (-1,1,2)

12) Given $ f(x) \equal{} x^5 \minus{} 5x^4 \plus{} 13x^3 \plus{} 121x^2 \minus{} 25x \plus{} 69$, find the sum of the cubes of its roots.

13) Tow poles are 50 feet apart on level ground. One pole is 10 feet taller than the other. A cable hangs form the top of the poles in the shape of a parabola. It's lowest point is 10 feet above a point on the ground 20 feet from the shorter pole. Find the height of the taller pole.

14) The phase shift of $ y \equal{} 5\sin3x \minus{} 12\cos3x$ is:

15)Solve for $ \theta$ where $ 0 < \theta < 90: \equal{} 2 \minus{} \sqrt3$

16) $ (\sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt {1 \plus{} ...}}})(\sqrt {(7 \minus{} 3\sqrt5) \plus{} \sqrt {(7 \minus{} 3\sqrt5) \plus{} \sqrt {(7 \minus{} 3\sqrt5) \plus{} ...}}}) \equal{}$

17)$ \sum_{x \equal{} 2}^{50}\frac {3}{(x \minus{} 1)(x \plus{} 1)}$
 \frac {\cos^2\theta - \sin^2\theta}{2\cos^2\theta\tan\theta}
18)How many values of x satisfy the equation: $ \log(x^2) \equal{} cos(x)$

19) Find the remainder when $ 19^{2001}$ is divided by 23.

Also, how would one go about simplifying something like : $ \sqrt {\sqrt2 \plus{} \sqrt5}$ or something like that to a single radical?

As soon as you get one, go ahead and post the solution.[/quote]

[hide="hint for 15"]$ \cos^2\theta \minus{} \sin^2\theta\equal{}\cos{2\theta}$

$ \cos^2\theta\tan\theta\equal{}\cos^2\theta\cdot\frac{\sin{\theta}}{\cos{\theta}}\equal{}\cos{\theta}\sin{\theta}\equal{}\sin{2\theta}$
 [/hide]

[hide="Hint for 6"]
The same as $ 1!\plus{}3!\plus{}5!$
[/hide]

[hide="Hint for 17"]
Multiply by $ \frac{\frac{3}{2}}{\frac{3}{2}}$
[/hide]

[hide="Hint for 18"]
Intersections of the graphs
[/hide]

[quote="ElChapin"][hide="Hint for 6"]
The same as $ 1! \plus{} 3! \plus{} 5!$
[/hide]

[hide="Hint for 17"]
Multiply by $ \frac {\frac {3}{2}}{\frac {3}{2}}$
[/hide]

[hide="Hint for 18"]
Intersections of the graphs
[/hide][/quote]

Please explain more for #17

[quote="sonny"]

[hide="hint for 15"]$ \cos^2\theta \minus{} \sin^2\theta \equal{} \cos{2\theta}$

$ \cos^2\theta\tan\theta \equal{} \cos^2\theta\cdot\frac {\sin{\theta}}{\cos{\theta}} \equal{} \cos{\theta}\sin{\theta} \equal{} \sin{2\theta}$
 [/hide][/quote]
I had this solved to $ \tan2\theta\equal{}2\minus{}\sqrt3$ but cannot figure it out from there.

[quote="brianb"]Hey yall here are some problems to solve:

Please explain and work fully:
(These are to be worked with no calculator, and in a method that can be feasibly used in a short amount of time, at least 2-3 minutes, preferably in 60 seconds or less)
Most will be somewhat formula based.

1) Find $ (.99)^{10}$ to the first three decimal places.

2)Find $ x^2 \plus{} y^2 \plus{} z^2 \equal{} a^2$ and $ x \plus{} y \plus{} z \equal{} x^3 \plus{} y^3 \plus{} z^3 \equal{} a$, write an equation expressing $ xyz$ in terms of $ a$.

3)Quadrilateral ABCD has right angles at B and D. If ABCD is kite shaped with AB-AD=20 and BC-CD=15, find the length of a radius of the circle inscribed in ABCD.

4) What kind of graph is $ r \equal{} 1 \plus{} 2\sin\theta$

5) Let S be the infinite sum of $ \dfrac{1}{2} \plus{} \dfrac{1}{2} \plus{} \dfrac{3}{8} \plus{} \dfrac{1}{4} \plus{} \dfrac{5}{32} \plus{} \dfrac{1}{8}...$
Let K be the sum of the squares of the roots of $ x^5 \minus{} 3x^4 \plus{} 6x^3 \minus{} 3x^2 \plus{} 1 \equal{} 0$
Find the product of S and K.

6) Find the remainder when $ 1! \plus{} 3! \plus{} 5!... \plus{} 49!$ is divided by $ 18$.

7)What is the $ 7th$ pentagonal number?

8) ABE is a right triangle with right angle E. C is a point on AE. D is a point on AB. CD forms a perpendicular line to AE. $ AC \equal{} 8, AD \minus{} 10, CD \minus{} 6,$ and  $ < BCD \equal{} 30 degrees.$ If BD= $ k(4\sqrt {3} \plus{} 3)$, find k.

9) The graph of the polar equation $ \cos\theta \plus{} r \equal{} \frac {1}{r}$ is a:

10)Evaluate: $ 2\arctan\frac {1}{3} \plus{} \arctan\frac {1}{7}$

11) Find the area of the triangle with vertices: (1,-3,-2) (5, -1, 2) (-1,1,2)

12) Given $ f(x) \equal{} x^5 \minus{} 5x^4 \plus{} 13x^3 \plus{} 121x^2 \minus{} 25x \plus{} 69$, find the sum of the cubes of its roots.

13) Tow poles are 50 feet apart on level ground. One pole is 10 feet taller than the other. A cable hangs form the top of the poles in the shape of a parabola. It's lowest point is 10 feet above a point on the ground 20 feet from the shorter pole. Find the height of the taller pole.

14) The phase shift of $ y \equal{} 5\sin3x \minus{} 12\cos3x$ is:

15)Solve for $ \theta$ where $ 0 < \theta < 90: \frac {\cos^2\theta \minus{} \sin^2\theta}{2\cos^2\theta\tan\theta} \equal{} 2 \minus{} \sqrt3$

16) $ (\sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt {1 \plus{} ...}}})(\sqrt {(7 \minus{} 3\sqrt5) \plus{} \sqrt {(7 \minus{} 3\sqrt5) \plus{} \sqrt {(7 \minus{} 3\sqrt5) \plus{} ...}}}) \equal{}$

17)$ \sum_{x \equal{} 2}^{50}\frac {3}{(x \minus{} 1)(x \plus{} 1)}$

18)How many values of x satisfy the equation: $ \log(x^2) \equal{} cos(x)$

19) Find the remainder when $ 19^{2001}$ is divided by 23.

Also, how would one go about simplifying something like : $ \sqrt {\sqrt2 \plus{} \sqrt5}$ or something like that to a single radical?

As soon as you get one, go ahead and post the solution.[/quote]

Here is the solution for number 19:
[hide="19"]
By Fermat's Theorem, $ 19^{22}\equiv 1\bmod 23$.  Thus, $ 19^{2001}\equiv [(19^{22})^{90})][19^{21}]\equiv (1)(19^{21})\equiv 19^{21}\bmod 23$.  Now, we say that $ 19^{21}\equiv k\bmod 23\implies 1\equiv 19^{22}\equiv 19k\bmod 23\implies 19k\equiv 1\bmod 23$.  Thus, $ \minus{} 19k\equiv \minus{} 1\bmod 23\implies 23k \minus{} 19k\equiv 23 \minus{} 1\bmod 23\implies 4k\equiv 22\bmod 23\implies 2k\equiv 11\bmod 23\implies 2k\equiv 23 \plus{} 11\equiv 34\bmod 23\implies \boxed{k\equiv 17\bmod 23}$.  [/hide]
Here are a few hints for the others:
[hide="Hint for 1"]
Binomially expand on $ (1 \minus{} .01)^{10}$.
[/hide]

[hide="Hint for 10"]
Set the value equal to $ x$ and apply the $ \tan$ function on both sides.
[/hide]

[hide="16"]
In general, we will find the value of $ \sqrt {x \plus{} \sqrt {x...}}$.  So, we call this value equal to y, and thus $ y \equal{} \sqrt {x \plus{} \sqrt {x...}}$.  Squaring both sides, we get that $ y^2 \equal{} x \plus{} \sqrt {x \plus{} \sqrt {x...}} \equal{} x \plus{} y\implies y^2 \minus{} y \minus{} x \equal{} 0$.
[/hide]

And why,oh why can't you just post it in only one forum?

[hide="2"]Expanding $ (x\plus{}y\plus{}x)^3$, we have
\[ (x\plus{}y\plus{}x)^3 \equal{} x^3\plus{}y^3\plus{}z^3\plus{}3(x^2y\plus{}y^2x\plus{}x^2z\plus{}z^2x\plus{}y^2z\plus{}z^2y)\plus{}6xyz\]
Multiplying x^2+y^2+z^2 and x+y+z, we have
\[ (x^2\plus{}y^2\plus{}z^2)(x\plus{}y\plus{}z)\equal{}a^3\]
\[ x^3\plus{}y^3\plus{}z^3\plus{}x^2y\plus{}y^2x\plus{}x^2z\plus{}z^2x\plus{}y^2z\plus{}z^2y\equal{}a^3\]
Let $ x^2y\plus{}y^2x\plus{}x^2z\plus{}z^2x\plus{}y^2z\plus{}z^2y\equal{}b$. We thus have the following two equations:
\[ a^3\equal{}a\plus{}3b\plus{}6xyz\]
\[ a\plus{}b\equal{}a^3\]
Solving for $ xyz$, we have $ xyz\equal{}\frac{\minus{}a^3\plus{}a}{3}$.[/hide]

[hide="hint for 17"]telescoping series[/hide]

[hide="hint for 5"]consider $ S\minus{}\frac12S$

I'd do some others but nah.[/hide]


$ \forall a,b,c > 0$ prove that

$ \frac {16a^2b^2}{\sqrt {2a^2b^2 \plus{} abc^2 \plus{} c^4}}\plus{}\frac {16b^2c^2}{\sqrt {2b^2c^2 \plus{} bca^2 \plus{} a^4}}\plus{}\frac {16c^2a^2}{\sqrt {2c^2a^2 \plus{} acb^2 \plus{} b^4}}\plus{}3(a^2\plus{}b^2\plus{}c^2)\ge 11(ab\plus{}bc\plus{}ca)$

[hide="note"] marry christmas :D  [/hide]

[quote="khanhsy"]$ \forall a,b,c > 0$ prove that

$ \frac {16a^2b^2}{\sqrt {2a^2b^2 \plus{} abc^2 \plus{} c^4}} \plus{} \frac {16b^2c^2}{\sqrt {2b^2c^2 \plus{} bca^2 \plus{} a^4}} \plus{} \frac {16c^2a^2}{\sqrt {2c^2a^2 \plus{} acb^2 \plus{} b^4}} \plus{} 3(a^2 \plus{} b^2 \plus{} c^2)\ge 11(ab \plus{} bc \plus{} ca)$

[hide="note"] marry christmas :D  [/hide][/quote]

marry christmas

$ \frac {16a^2b^2}{\sqrt {2a^2b^2 \plus{} abc^2 \plus{} c^4}} \plus{} \frac {16b^2c^2}{\sqrt {2b^2c^2 \plus{} bca^2 \plus{} a^4}} \plus{}$ $ \frac {16c^2a^2}{\sqrt {2c^2a^2 \plus{} acb^2 \plus{} b^4}}$

$ \geq \frac {16(\sum a^2b^2)^2}{\sum (a^2b^2\sqrt {2a^2b^2 \plus{} abc^2 \plus{} c^4})} \equal{}$ $ \frac {16(\sum a^2b^2)^2}{\sum (ab\sqrt {a^2b^2(2a^2b^2 \plus{} abc^2 \plus{} c^4)})}$

$ \geq \frac {16(\sum a^2b^2)^2}{\sqrt {a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}\sqrt {\sum {(a^2b^2(2a^2b^2 \plus{} abc^2 \plus{} c^4))}}}$

we only prove

$ \frac {16(\sum a^2b^2)^2}{\sqrt {a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}\sqrt {\sum {(a^2b^2(2a^2b^2 \plus{} abc^2 \plus{} c^4))}}}\geq 8(ab \plus{} bc \plus{} ca)$

$ \Longleftrightarrow \frac {4(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)^3}{\sum {(a^2b^2(2a^2b^2 \plus{} abc^2 \plus{} c^4))}}\geq (ab \plus{} bc \plus{} ca)^2$

The inequality holds,but i can not prove.


function
logarithms
integration
limit
calculus
calculus computations

How does it prove that

$ e^x  \equal{} \sum\limits_{k \equal{} 0}^\infty  {\frac{{x^k }}{{k!}}} .$

[quote="Hetidemek"]How does it prove ...[/quote]

That depends on what it is, and what its definition of the function $ e^x$ is.  (In English, the construction you're looking for is, "How does one prove ... ?"  "It" isn't used to refer to people, except derogatorially.)

by the way, taking the above as the definition of $ e^x$ is a perfectly good definition.

Alternate definitions of $ e^x$ include

- the unique function $ \exp(x)$ such that $ \frac{d}{dx} \exp(x) \equal{} \exp(x)$ and $ \exp(0) \equal{} 1$.
- the inverse of the function $ \ln x \equal{} \int_{1}^{x} \frac{dt}{t}$.
- $ \lim_{n \to \infty} \left( 1 \plus{} \frac{x}{n} \right)^n$.

The thing to do, then, is to prove that all of the definitions are equivalent, which isn't so bad.  (In analysis, I understand we usually take the series definition.)


inequalities
function
inequalities unsolved

you did nothing. you proved ${\frac{\left(a^{4}+a^{2}+1\right) \left(b^{4}+b^{2}+1\right)}{\left(a^{4}+a^{2}\right) \left(b^{4}+b^{2}\right)}\geq \left(\frac{a^{2}b^{2}+a b}{a^{2}b^{2}+a b}\right)^{2}}$, not ${\frac{\left(a^{4}+a^{2}+1\right) \left(b^{4}+b^{2}+1\right)}{\left(a^{4}+a^{2}\right) \left(b^{4}+b^{2}\right)}\geq \left(\frac{a^{2}b^{2}+a b+1}{a^{2}b^{2}+a b}\right)^{2}}$. and you don't understand behemont's statement. he meant you just proved $A \geq C$ where the problem is asking to prove $A \geq B$ where $B \geq C$. you gotta substitute $A={\frac{\left(a^{4}+a^{2}+1\right) \left(b^{4}+b^{2}+1\right)}{\left(a^{4}+a^{2}\right) \left(b^{4}+b^{2}\right)}}$, ${B=\left(\frac{a^{2}b^{2}+a b+1}{a^{2}b^{2}+a b}\right)^{2}}$, and ${C=\left(\frac{a^{2}b^{2}+a b}{a^{2}b^{2}+a b}\right)^{2}}$.

anyhow, the original problem can be solved by using Jensen's on a function $f(x)=ln(1+\frac{1}{e^{2x}+e^{x}})$.

Sung-yoon Kim can you complete proof? Thanks.

sure.

$f(x)=ln(1+\frac{1}{e^{2x}+e^{x}}$
$f'(x)=-\frac{2e^{x}+1}{(e^{x}+1)(e^{2x}+e^{x}+1}$
the numerator of $f''(x)$ is $(2e^{x}+1)(3e^{3x}+4e^{2x}+2e^{x})-2e^{x}(e^{3x}+2e^{2x}+2e^{x}+1)$
$=4e^{4x}+7e^{3x}+4e^{2x}>0$.
hence, the function is convex, so that we can apply the Jensen's.

now let $a=e^{x/2},b=e^{y/2}$. then the original inequality is equivalent to $f(x)+f(y) \geq 2f(\frac{x+y}{2})$, which is true by Jensen's.

Thank you Sung-yoon Kim. It was a good job.  :roll:

it can be solved if one multiplies everything, and then AM-GM.

but this one is much prettier


induction
geometry
geometric transformation
rotation
geometry unsolved

Let ABCD be a square.
1)Prove that for any $ n$ greater or equal with 3, the square ABCD can be partitoned in [b]two congruent polygons with n sides[/b]
2)Prove that for any $ n$ greater or equal with 3, the square ABCD can be partitoned in [b]four congruent polygons with n sides[/b]

Let ABC be an equilateral triangle.

[b]3)Prove that it can be partitioned in $ n^2$ equilateral triangles,for every $ n > \equal{} 1$
4)Prove that it can be partitioned in 3 congruent polygons with $ n$ sides,for every $ n > \equal{} 3$.[/b]

3,It is useful, here is case $ n \equal{} 4$.

Generalization: Every regular $ k$-gon can be partition in $ k$ equal polygons with $ n (n\ge k)$ sides.

[hide="hint"]Consider the center of the regular $ n$-gon and rotations of $ 2 \Pi /n$ with that center[/hide]

could you go into more detail?for example,the case of the equilateral triangle

4) Case for an equilateral triangle initial case for $ n\equal{}3$
Let $ O$ be the center of the triangle then $ AOB$ , $ BOC$ & $ COA$ is the desired partition.
Consider a polygonal $ C_1C_{2}\cdots C_k$ inside $ AOB$ then
rotate it with center $ O$ and angle $ 2 \Pi /3$ obtaining $ A_1A_{2} \cdots A_k$ and $ B_1B_{2} \cdots B_k$
Notice that the equal polygons
$ AC_1C_{2} \cdots C_kOA_{k} \cdots A_1B$,
$ BA_1A_2 \cdots A_kOB_k \cdots B_1C$ and
$ CB_1B_2 \cdots B_kOC_k \cdots C_1A$ have $ n\equal{}2k\plus{}3$ sides.
For the case of even $ n$ start with an initial case of $ n\equal{}4$
Then generalize for regular $ k$-gon


I know there is a formula for the sum of the first n squares, but i don't rembmer it.  Does anyone here remember it?

The formula for the first n squares is:
n(n+1)(2n+1)/6
Since we're on the topic, if you were wondering about the formula for the sum of the first n cubes, it's:
(n :^2: *(n+1) :^2: )/4

o yeah. you're right, that's probably much easier. i'll remember to use that formula next time.

How those n square formulas work?
Can anyone tell me how that thing does work?
It could be really useful...


Miko always remembers his four-digit PIN (personal identification number) because
(a) it is a perect square, and
(b) it has the property that, when it is divided by 2, or 3, or 4, or 5, or 6, or 7, or 8, or 9, there is always a remainder of 1.
What is Miko's PIN?

Let the number be $ n$.
Then $ lcm(2,3,4,5,6,7,8,9)\equal{}2520|n\minus{}1$
So the possible values of $ n$ are $ 2521, 5041$ and $ 7561$
After checking, $ 5041\equal{}71^2$ is the answer.


[url=http://www.arml.com/Contest%20Information/Song%20Contest.pdf]Contest rules (PDF)[/url]

I tried to get a team together last year, but it fell apart. We need to enter this year.

Fact: Last year, there were fewer teams in the contest than were spaces available. We could have passed the qualifying round with me screeching alone on the stage.

Fact: Every year, most of the teams are awful. There is always at least one team that gets more boos than cheers. If we compete, they'll be struck silent by our avant-garde disregard for the status quo. No booing, and no annoying cheering.

We need maybe 5-10 people. Our current idea is to wobble around like maniacs and say "Meeg."

I'd like to do the song contest as well. If I'd known you were trying to last year, I would have definitely joined in.
Welcome to the forums, by the way.

I'm willing to do it...we just need to get organized...

ah yes, i remember the abortive attempt at "meeg in E minor"


inequalities
algorithm
linear algebra
matrix

Prove that for non-negative a, b, c

1.
$ \frac{a\plus{}\sqrt{ab}}{2} \leq \sqrt{a \times \frac{a\plus{}b}{2}}$

2.
$ \frac{a\plus{}\sqrt{ab}\plus{}\sqrt[3]{abc}}{3} \leq \sqrt[3] { a \times \frac{a\plus{}b}{2} \times \frac{a\plus{}b\plus{}c}{3} }$

Does this generalise?

[hide="1"]Square both sides:
\[ \frac{a^2\plus{}ab\plus{}2a\sqrt{ab}}{4}\leq \frac{ab\plus{}a^2}{2}\]
Get rid of the fractions:
\[ 4ab\plus{}4a^2\geq 2a^2\plus{}2ab\plus{}4a\sqrt{ab}\]
Move to one side:
\[ 2ab\plus{}2a^2\minus{}4a\sqrt{ab}\geq 0\]
Divide by $ 2a$:
\[ b\plus{}a\minus{}2\sqrt{ab}\geq 0\]
And by AM-GM that is true.[/hide]

[hide="1 2 and Generalization"]In the general case, apply Holder with equal exponents to an $ n! \times n!$ matrix. I'll give the matrices for $ n \equal{} 2,3$. I haven't quite nailed down the algorithm for determining it for arbitrary $ n$, but I'm pretty certain it exists. The idea is to define $ (n \minus{} 1)! \times (n \minus{} 1)!$ submatrices with equal numbers of variables in each row and column.

$ n \equal{} 2$:
$ \begin{bmatrix} \sqrt {a/2} & \sqrt {a/2} \\
\sqrt {b/2} & \sqrt {a/2} \end{bmatrix}$

$ n \equal{} 3$:
$ \begin{bmatrix} \sqrt [6]{\frac {a}{6}} & \sqrt [6]{\frac {a}{6}} & \sqrt [6]{\frac {a}{6}} & \sqrt [6]{\frac {a}{6}} & \sqrt [6]{\frac {a}{6}} & \sqrt [6]{\frac {a}{6}} \\
\sqrt [6]{\frac {a}{6}} & \sqrt [6]{\frac {a}{6}} & \sqrt [6]{\frac {a}{6}} & \sqrt [6]{\frac {a}{6}} & \sqrt [6]{\frac {a}{6}} & \sqrt [6]{\frac {a}{6}} \\
\sqrt [6]{\frac {b}{6}} & \sqrt [6]{\frac {b}{6}} & \sqrt [6]{\frac {b}{6}} & \sqrt [6]{\frac {a}{6}} & \sqrt [6]{\frac {a}{6}} & \sqrt [6]{\frac {a}{6}} \\
\sqrt [6]{\frac {b}{6}} & \sqrt [6]{\frac {b}{6}} & \sqrt [6]{\frac {a}{6}} & \sqrt [6]{\frac {b}{6}} & \sqrt [6]{\frac {a}{6}} & \sqrt [6]{\frac {a}{6}} \\
\sqrt [6]{\frac {c}{6}} & \sqrt [6]{\frac {c}{6}} & \sqrt [6]{\frac {b}{6}} & \sqrt [6]{\frac {b}{6}} & \sqrt [6]{\frac {a}{6}} & \sqrt [6]{\frac {a}{6}} \\
\sqrt [6]{\frac {c}{6}} & \sqrt [6]{\frac {c}{6}} & \sqrt [6]{\frac {b}{6}} & \sqrt [6]{\frac {b}{6}} & \sqrt [6]{\frac {a}{6}} & \sqrt [6]{\frac {a}{6}} \\
\end{bmatrix}$
[/hide]

Thanks, I proved 1 using

[hide]Jensen on $ f(x) \equal{} \sqrt{x}$ and $ \{ x^2, xy\}$[/hide]

But I couldn't do 2

2 by Holder (from Winning Solutions by Lozansky)

$ (a\plus{}a\plus{}a)^{\frac{1}{3}} (a\plus{}\sqrt{ab}\plus{}b)^{\frac{1}{3}} (a\plus{}b\plus{}c)^{\frac{1}{3}} \ge a\plus{}\sqrt{ab}\plus{}\sqrt [3] {abc}$

Further $ a\plus{}b\plus{}\sqrt {ab} \le a\plus{}b\plus{}\frac{a\plus{}b}{2} \equal{} \frac{3(a\plus{}b)}{2}$

This leads to the given inequality

The first one is also similarly from Cauchy Schwarz (or Holder) $ a\plus{}\sqrt{ab} \le \sqrt{(a\plus{}a) (a\plus{}b)}$

Anyone has idea of generalisation for n variables?

This is the [url="http://www.dma.ulpgc.es/profesores/personal/aph/ficheros/resolver/ficheros/crp/mixed_1994.pdf"]Mixed AM-GM Inequality.[/url]


trigonometry
inequalities
function
inequalities unsolved

[quote="nguyenvanhuyen"]Given $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} abc$. Prove that:
$ \frac {1}{\sqrt {a^2 \plus{} 1}} \plus{} \frac {1}{\sqrt {b^2 \plus{} 1}} \plus{} \frac {1}{\sqrt {c^2 \plus{} 1}} \le \frac {3}{2}$[/quote]

Ineq
$ <\equal{}> \sum \sqrt{\frac{bc}{(a\plus{}b)(a\plus{}c)}} \le \frac{3}{2}$
It's true by Am-Gm  :oops:

use trig: sub in a=tan x, b=tan y, c=tan z, (x+y+z=pi, 0<x,y,z<pi/2) which satisfies the condition a + b + c = abc
then RTP cos x + cos y + cos z <= 3/2
which holds by Jensen

[quote="raveen.ds"]use trig: sub in a=tan x, b=tan y, c=tan z, (x+y+z=pi, 0<x,y,z<pi/2)[/quote]
How do you know that $ x\plus{}y\plus{}z\equal{}\pi$?

from a + b + c = abc,
c = (a+b) / (ab-1)
letting a = tan x, b = tan y, then c = - tan (x+y) = tan (pi-x-y)
so x + y + z = pi

[hide="my solution"]

By AM-GM we have $ a\plus{}b\plus{}c \geq 3 \sqrt[3] {abc}$ and since $ a\plus{}b\plus{}c\equal{}abc \implies (abc)^2 \geq 27$

We rewrite the given inequality as

$ \frac{1}{3} \left(\frac{1}{a^2 \plus{}1} \plus{}\frac{1}{b^2 \plus{}1} \plus{}\frac{1}{c^2 \plus{}1} \right) \leq \frac{1}{2}$

Since  function $ f(a)\equal{}\frac{1}{\sqrt{a^2 \plus{}1}}$ is concave ,we apply Jensen's inequality

$ \frac{1}{3}f(a) \plus{}\frac{1}{3}f(b) \plus{}\frac{1}{3}f(c) \leq f\left(\frac{a\plus{}b\plus{}c}{3} \right)\equal{}f \left(\frac{abc}{3}\right)\equal{}\frac{1}{\sqrt{\frac{(abc)^2}{3^2} \plus{}1}}\leq \frac{1}{2} <\equal{}> (abc)^2 \geq 27, \mathbb{QED}$ [/hide]

[hide="@caubetoanhoc94"]Can you post your full solution caubetoanhoc94  :blush: [/hide]

[quote="enndb0x"][hide="my solution"]

By AM-GM we have $ a \plus{} b \plus{} c \geq 3 \sqrt [3] {abc}$ and since $ a \plus{} b \plus{} c \equal{} abc \implies (abc)^2 \geq 27$

We rewrite the given inequality as

$ \frac {1}{3} \left(\frac {1}{a^2 \plus{} 1} \plus{} \frac {1}{b^2 \plus{} 1} \plus{} \frac {1}{c^2 \plus{} 1} \right) \leq \frac {1}{2}$

Since  function $ f(a) \equal{} \frac {1}{\sqrt {a^2 \plus{} 1}}$ is concave ,we apply Jensen's inequality

$ \frac {1}{3}f(a) \plus{} \frac {1}{3}f(b) \plus{} \frac {1}{3}f(c) \leq f\left(\frac {a \plus{} b \plus{} c}{3} \right) \equal{} f \left(\frac {abc}{3}\right) \equal{} \frac {1}{\sqrt {\frac {(abc)^2}{3^2} \plus{} 1}}\leq \frac {1}{2} < \equal{} > (abc)^2 \geq 27, \mathbb{QED}$ [/hide]

[hide="@caubetoanhoc94"]Can you post your full solution caubetoanhoc94  :blush: [/hide][/quote]

Yes. :P
$ 2\sqrt {\frac {bc}{(a \plus{} b)(a \plus{} c)}} \le  \frac {b}{a \plus{} b} \plus{} \frac {c}{a \plus{} c}$


Find all triples $ (a,b,c)$ of positive integers such that

$ (1\plus{}\frac{1}{a})(1\plus{}\frac{1}{b})(1\plus{}\frac{1}{c})\equal{}2$

[quote="Sashsiam_2"]Find all triples $ (a,b,c)$ of positive integers such that

$ (1 \plus{} \frac {1}{a})(1 \plus{} \frac {1}{b})(1 \plus{} \frac {1}{c}) \equal{} 2$[/quote]

hello,at first a<b<c
 case 1) a=1 and there are no solutions
 case 2) a=2 and then we get 3 solutions
 case 3) a=3 also one solution
now when a=> 4 then b=>5 ,c=>6 and we see that there are no solutions.
so i found just,
(a,b,c)-->(2;4;15),(2;5;9),(2;6;7),(3;4;5).

http://www.kaynet.or.jp/~kay/misc/den2.html
[url=http://www.research.att.com/~njas/sequences/A085098]A085098[/url] - OEIS

[b]Program of Mathematica[/b]
Timing[For[a = 1, a ≤ 15, a++, {For[b = a, b ≤ 15, b++, {For[c = b, c ≤ 15, c++, If[(1 + 1/a) (1 + 1/b) (1 + 1/c) == 2, Print[{a, b, c}]]]}]}]]

[b]Answer and Time[/b]
{2,4,15}
{2,5,9}
{2,6,7}
{3,3,8}
{3,4,5}

{0.01 Second,Null}


There are $n$ chair's, $1,2,...,n$ around a table. $n$ people, $1,2,...,n$ sit on the chairs (in some order).
a) If $n$ is even, prove that there is $k\geq 0$ such that if every person moves for $k$ places in the same direction there is no person which has the same number as the chair it sits on;
b) If $n$ is odd prove that there is starting arrange such that there is no such $k$.

for even $n$. we get for every $k=1,2,3,4...,n$, there is a guy who falls in the seat with his number....so for every k there is exactly one guy!....so we get that person $i$ is in the seat $i+a_i$ mod n, where $a_i$ is a permutation of $1,2,3...,n$. so we get suming all positions that $0=n(n+1)/2$ modulo $n$, contradiction.

Now for odd $n$, we put person $i$ in seat $2i$, and it works...


The Indian team to the IMO 2006 is :

1) Abhishek Dang - Pune (Maharashtra)
2) Varun Jog - Pune (Maharashtra)
3) Riddhipratim Basu - West Bengal
4) Apurv Nakade - Nagpur (Maharashtra)
5) Ashay Burungale - Satara (Maharashtra)
6) Manas Joglekar - Mumbai (Maharashtra)

[color=blue][size=200]BEST OF LUCK TO THE TEAM![/size][/color]

Riddhi says India won six gold medals each time from '98 to '04, so there is little reason not to hope it will be the same this year. But it is good to see Vihang back.

Best of luck Riddhi, Ashay, Manas and all the rest!

riddhi is not mad enough to say so

india has never won more than 3 golds
 riddhi said  that india won 6 medals ,not golds


function
limit
calculus
calculus computations

The C.D.F. (cumulative distribution function) of the random variable X is given by,

F(x)=[0,  x<0
         x/2,  0<= x<1
         2/3,  1<= x<2
         11/12,  2<= x<3
         1,  3<=x ]

a) Find P(X>1/2)
b) Find P(2<X<=4)
c) Find P(X=1)

Does anyone know how to answer this question?

It wouldn't hurt to sketch the graph of this function.

The definition of cumulative distribution function:

$F(x)=P(X\le x).$

This forces the c.d.f. to be continuous from the right. As for limits from the left,

$\lim_{x\to a^{-}}F(x)=P(X<a).$

Let's use the shorthand notation $F(a^{-})$ to mean $\lim_{x\to a^{-}}F(x).$

Here's a table of what we know:

$P(X<a)=F(a^{-})$
$P(X\le a)=F(a)$
$P(X>a)=1-F(a)$
$P(X\ge a)=1-F(a^{-})$
$P(a<X<b)=F(b^{-})-F(a)$
$P(a<X\le b)=F(b)-F(a)$
$P(a\le X<b)=F(b^{-})-F(a^{-})$
$P(a\le X \le b)=F(b)-F(a^{-})$
$P(X=a)=F(a)-F(a^{-})$

So, your questions:

$P\left(X>\frac12\right)=1-F\left(\frac12\right)=\frac14$

$P(2<X\le 4)=F(4)-F(2)=1-\frac{11}{12}=\frac1{12}$

$P(X=1)=F(1)-F(1^{-})=\frac23-\frac12=\frac16$


Hello.

Here are a few problems I couldnt solve or at least not easily. The idea is making euclidean constructions with ruler and compass. They all refer to hyperbola $ H$.

Thanks for your suggestions.

(A) Find foci F and F' given the vertexes and a point $ P\in H$.

[hide]We know $ 2a \equal{} VV'$ and so we can find $ 2c \equal{} FF'$ in terms of $ a$ and the distances from $ P$ to axes.
I cant find a simple metric construction to find FF'[/hide]

(B) A focus F, two tangentes and one point are given. Find F', the other focus.

[hide] Couldnt solve it. One of my ideas is the following.
Let $ t_1$ and $ t_2$ be the tangent lines and $ M$ the given point. The symmetric of the focus F with respect to $ t_1$ and $ t_2$ belong to the circle with center F' and radius $ 2a$. Now it would be fine a relationship between $ M$ and the circle with center $ F'$.
[/hide]

(C) A focus, one tangent and two points are given. (F, t, M and P resp.) Find F', the other focus.

[hide]I solved it but one part of my solution is difficult. F' is the center of the circle tangent to the circles M(F) and N(F) and passes thru the symmetric of F with respect to the tangent. This circle is not easy to find without inversion?.[/hide]

(D) A focus (F) and three points (A,B,C) are given.

[hide]This case reduces to the Apollonius problem:Find a circle centered at F' that is tangent to the circles A(F), B(F) and C(F). Maybe there is a simpler solution than in the general case because all the circles pass thru the same point F., and avoid inversion.[/hide]

Any suggestions?

I had no replies...  :(

Do you think I should move it to another Forum? Which one? :huh:

I'm a stupid. Solution to (B) is easy. F' is the center of the circle tangent to $ A(F)$ and passes thru $ F_1$ and $ F_2$ symmetrics of F with respect to tangents.
This case is then PPC (Apollonius problem again).

Now, I solved all problems but (A) which is pending of solutions... The others are solved but I think some must have a simpler solution than mine, esp. (C) and (D).


and I think it is also china TST.
$abcd=1$
then ;
$\frac 1 {(1+a)^2}+\frac 1 {(1+b)^2}+\frac 1 {(1+c)^2}+\frac 1 {(1+d)^2}\ge 1$
let $d=1$ get you want :)

OK, zhaobin. Present for you from me:
$a>0,b>0,c>0.$ Prove that $\sum(\frac{a}{a+b})^{1.6}\geq\frac{3}{2^{1.6}}.$ :lol:

[quote=zhaobin]a little bigger present  :P
$xyz=1,x,y,z \ge 0$
then $\frac 1 {(1+x)^k}+\frac 1 {(1+y)^k}+\frac 1 {(1+z)^k} \ge \min \{ 1,\frac 3 {2^k} \}$ :D[/quote]

Two Generalization:
[b]The First[/b] Let $\[{x_1},{x_2}, \cdots ,{x_n} \ge 0\]$ be such that their product is 1,then
\[\frac{1}{{{{\left( {1 + {x_1}} \right)}^k}}} + \frac{1}{{{{\left( {1 + {x_2}} \right)}^k}}} +  \cdots  + \frac{1}{{{{\left( {1 + {x_n}} \right)}^k}}} \ge \min \left\{ {1,\frac{n}{{{2^k}}}} \right\}.\]
[b]The Second[/b] [url=https://www.artofproblemsolving.com/community/q1h1383969p7704261]Proved by the RCF Theorem[/url]


As the old Games & Fun Factory Marathon started by 236Factorial kind of died out, i am creating a new one.

You can read the rules here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47997[/url]

Question 1: In the Star Wars Saga, whom does Samuel L. Jackson play?

Samuel Jackson plays Mace Windu ,
senior member of the jedi council

Question 2: What is the name of the girlfriend of Eragon's brother in the Inheritance trilogy?

Question 3 : What is the real name of Storm
                      from the series X-men?

what gave you that idea?
  i meant the real world name .
you know, like Superman was Clarke Kent,
Spiderman was Peter Parker etc.

  By the way, kheg and shadysurspammed , you
are perfectly qualified to join the annual avatar showcase on the
Fun Factory. Check it out.

Question 4: What is the title of Dick Feynman's bestselling autobiography?


Is there any way to easily compute a large combination like $\dbinom{132}{56}$ without using a calculator or writing out (132*131*130*129* ... ) / (56*55*54*53* ... )? Could anybody enlighten me?

Not really. Like the answer to your example is $83415033495866276053771034616980883120$. They tend to not be too fun to calculate.

But, of course, you can write out the terms and cancel a bunch of factors.


integration
calculus
calculus computations

[size=150][color=red]$ \int_{x \equal{} 1}^{2}\int_{y \equal{} \sqrt {x}}^{x} sin(\frac {x \pi}{2y})dydx \plus{} \int_{x \equal{} 2}^{4}\int_{y \equal{} \sqrt {x}}^{2} sin(\frac {x \pi}{2y})dydx \equal{} \frac {4 \pi \plus{} 8}{ \\
pi^3 }$[/color][/size]


trigonometry
vector
abstract algebra
complex numbers

[hide="Proof"]Let $r=|a|=|b|=|c|=|z|$. If we put $a=re^{i\alpha}, b=re^{i\beta}, c=re^{i\gamma}, z=re^{i\theta}$, we get:

$\cos\alpha+\cos\beta+\cos\gamma=\cos\theta$
$\sin\alpha+\sin\beta+\sin\gamma=\sin\theta$

Squaring both equations and adding them up, we get:

$3+2(\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha))=1$
$\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)=-1$
$1+\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)=0$

Using half-angle formulas and sum of cosines formula, we get:

$2\cos^2{{\alpha-\beta} \over 2}+2\cos{{\beta-\alpha} \over 2}\cos{{\alpha+\beta-2\gamma} \over 2}=0$
$cos{{\alpha-\beta} \over 2}(\cos{{\alpha-\beta} \over 2}+\cos{{\alpha+\beta-2\gamma} \over 2})=0$

Using sum of cosines formula again, we get:

$cos{{\alpha-\beta} \over 2}\cos{{\beta-\gamma} \over 2}\cos{{\gamma-\alpha} \over 2}=0$      (*)

This means that at least one of the following is true: $\alpha-\beta=\pm\pi$, $\beta-\gamma=\pm\pi$, $\gamma-\alpha=\pm\pi$. Let's say $\alpha=\beta\pm\pi$. That gives $e^{i\alpha}=-e^{i\beta}$ i.e. $a=-b$, giving $c=z$. If two of the factors in (*) are equal to zero, it's easy to show that exactly two of $a,b,c$ are equal to $z$ and the remaining is equal to $-z$. Not all of the factors can be equal to zero, because if $\alpha-\beta=\pm\pi$ and $\beta-\gamma=\pm\pi$, adding those gives $(\gamma-\alpha)\in\lbrace{0,\pm 2\pi\rbrace}\Rightarrow\cos{{\gamma-\alpha} \over 2}=\pm 1$.[/hide]

[hide]
How about this:

Connect $a$, $b$, and $c$ like you're adding complex numbers (in vector form) starting with $a$ at the origin in the complex plane. You end up at the complex number $z$. But since they all have equal magnitudes, the shape formed is a rhombus. Hence $z$ is parallel to one of the three. Therefore it has equal magnitude and angle.[/hide]

Nice, I was looking for a simple vector solution like that  :)

[quote="silouan"]Let  $a,b,c,z\in C$  with  equal    lenghts  and  $a+b+c=z$
Prove that at least one from the $a,b,c$ is equal to  $z$[/quote]

I came across this nice problem last year in a journal (if i remember exactly)..

I solved it like this...:

Since they are of the same module,say $\mid a \mid= \mid b \mid= \mid c \mid= \mid z \mid=r \in \mathbb{R}$
Thus $a\overline{a}=b\overline{b}=c\overline{c}=r^2$

Note that
$\overline{a}+\overline{b}+\overline{c}=\overline{z}$ or, due to the above,

$\frac{r^2}{a}+\frac{r^2}{b}+\frac{r^2}{c}=\frac{r^2}{z}$ or

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{z}=\frac{1}{a+b+c}$

or, doing some easy calculations

$(a+b)(b+c)(c+a)=0$ and the conclusion foolows   :D  :)


function
integration
calculus
Functional Analysis
topology
advanced fields
advanced fields unsolved

$E=C([0;1], \mathbb{R})$ the Banach space of continuous function on $[0;1]$ with the sup norm.
$S$ the subset of $f \in E$ that satisfy $\int_0^1 f(x)^2 dx = 1$. Is $0$ in the closure of $S$ for the weak topology.

I believe the answer is positive.

A functional $\mu$ on $C(I)$ ($I=[0,1]$) is a Borel measure on $I$ (which we'll denote by $\mu$ as well) such that $\mu(f)=\int_If\mbox d\mu,\ \forall f\in C(I)$. A weakly open subset of $C(I)$ contained in the standard local basis around $0$ is a set of the form $V_\varepsilon(\{\mu_i\}_{i=1}^n)=\{f\in C(I)\ |\ |\mu_i(f)|<\varepsilon,\ i=\overline{1,n}\}$.

Now fix such an open set $V_\varepsilon(\mu_i)$. We want to prove that there are functions $f\in V_\varepsilon(\mu_i)$ such that $\int_If^2\mbox dt=1$. I'd like to do this: first choose an open subinterval $U$ of $I$ with $|\mu_i|(U)$ small enough for what follows. Then to define $f$ to be $0$ outside $U$, with $\sup_I f=M$, and very close to the constant function $M$ in the $L^2$ norm on $U$. $M$ should satisfy the following conditions: on the one hand $M\cdot|\mu_i|(U)<\varepsilon,\ i=\overline{1,n}\ (*)$, which will ensure $f\in V_\varepsilon(\mu_i)$, and then $M^2\ell(U)\ge 2\ (**)$, where $\ell(U)$ is the length of $U$; this will ensure $\int_If^2\mbox d\mu>1$, after which we can scale $f$ down to a function with integral of the square precisely $1$, which will, of course, still belong to $V_\varepsilon(\mu_i)$.

From $(*),(**)$ we see that we want an $M$ such that $\frac 2{\sqrt{\ell(U)}}\le M<\frac\varepsilon{|\mu_i|(U)},\ i=\overline{1,n}$. It will be possible to find such an $M$ iff $\forall i\in\overline{1,n},\ \frac 2\varepsilon|\mu_i|(U)<\sqrt{\ell(U)}\ (1)$. Now, if we define the measure $\mu$ by $\mu=\frac 2\varepsilon\sum_{i=1}^n|\mu_i|$, $(1)$ is automatically satisfied if we choose $U$ such that $\mu(U)<\sqrt{\ell(U)}\ (2)$. Note that given any positive measure $\mu$ on $I$ we can always choose such an interval $U$: first divide $I$ into $2\ \ell$-equal parts, and choose the half with smaller $\mu$. Then $\mu$ has decreased twice, whereas $\sqrt{\ell}$ has decreased $\sqrt 2$ times. Now for the chosen interval repeat the procedure, and so on, until we reach an interval $U$ which satisfies $(2)$.


http://web.mit.edu/~tmildorf/www/Inequalities.pdf

Page 23/24.


calculus
integration
symmetry
algebra
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search
number theory

Let a,b, and c belong to the set of positive and negative integers.

Is there a solution to the following equation?

a/b + b/c + c/a = 1

a/b + b/c + c/a = 1

a, b, c are positive or negative integers. We have got 4 posibilities.

(i) All the three numbers are positive integers.
(ii) All the three numbers are negative integers.

For the above two combinations the equations gets transformed in
a/b + b/c + c/a = 1 where the integrals a, b, c > 0.      ......(1)

(iii) Two integers are positive and one is negative.
(iv) Two integers are negative and one is positive.

If we consider the symmetry of the equation, it transforms into
a/b - b/c - c/a = 1 where the integrals a, b, c > 0.         ......(2)

Case (1):
a/b + b/c + c/a = 1 where [size=150][b]the integrals a, b, c > 0[/b][/size].
If this is the case then a, b, or c cannot be equal to any other and the summation would be greater than 1. And, when they are unequal then at least any one of the terms a/b, b/c and c/a > 1 and thus, the equation cannot hold true. So we are left with the second case.

Case (2):
a/b - b/c - c/a = 1 where [b][size=150]the integrals a, b, c > 0[/size][/b].
It implies that a > b as a/b > 1. Say, a = nb where n > 1.

If b = c, then a/b - b/a = 2.  So, a = (1 + sqrt2)b which doesn't permit integral value to a. Therefore, b is not equal to c.

I am stuck at this point.

[quote="Shamik Banerjee"]
I am stuck at this point[/quote]
I got that far too. I see no way to continue, yet. :(

Wheres Andrew Wiles when you need him? :lol:

[quote="Shamik Banerjee"]This is not that easy as I thought earlier and proceeded :(

a/b + b/c + c/a = 1
=> a^2*c + b^2*a + c^2*b = abc, which is a nonsingular homogeneous cubic polynomial equation, also known as an elliptic curve, although not in standard (Weierstrass) form.

You can put it in Weierstrass form using the birational transformation
  a = 8x
  b = 4(x + y + 4)
  c = -x(x + y + 4)

With the reverse transformation
  x = -4b/c
  y = 4(a^3*c - a^3*b - b^2*c^2)/(a^3*b)

(where we want any rational solutions in x and y), and the new equation is
  y^2 = x^3 + x^2 + 8x + 16.

After some analysis on this elliptic curve, and we find that it has rank zero and only three rational points, namely   (0, 4),  (0, -4), and the "point at infinity"; (0, 1, 0) in homogeneous coordinates.

Those three points map back to the three (a, b, c) solutions
  (0, 1, 0),
  (1, 0, 0),
  (0, 0, 1),

to the equation   a^2*c + b^2*a + c^2*b = abc, respectively, but of course all three of these solutions are not solutions to the original equation   a/b + b/c + c/a = 1.

So we conclude that there are no integer or rational solutions to the equation.

  http://www.math.mcgill.ca/connell/
  http://www.math.mcgill.ca/connell/public/ECH1/
 Look in chapter one.

  Cubic Diophantine Equation in Three Variables
  http://mathforum.org/library/drmath/view/66650.html   and you can also search for "elliptic curve."

http://pari.math.u-bordeaux.fr/

That's all I can say  :D[/quote]

I can't even understand half the words you're using. (That's my way of saying that I'm ignorant.) Shouldn't there be a simpler way?

This problem is evil. It was posted here a lonnng time ago. We searched for an answer to no avail so I posted it on the alt.math forum. I was given the same answer that Shamik Banerjee found - namely that it reduces to a problem of finding rational points on an elliptic curve. I don't think it has a more elementary solution.


EDIT: I know the answer...
EDIT2: WOW! You guys are hysterical.

I bet a lot of you know this problem but I think it's pretty fun.
Also, if it's been said before I am sorry. Don't call me a spammer because I am sharing a fun problem.

A rope over the top of a pulley has the same length on each side. It weighs one third of a pound per foot. On one end hangs a monkey holding a banana, and on the other end is attached a weight equal to the weight of the monkey and the banana. The banana weighs two ounces per inch. The rope is as long (in feet) as the age of the monkey (in years), and the weight of the monkey (in ounces) is the same as the age of the monkey's mother.

The combined ages of the monkey and its mother are thirty years. One half the weight of the monkey, plus the weight of the banana, is one fourth as much as the weight of the weight and the weight of the rope. The monkey's mother is half as old as the monkey will be when it is three times as old as its mother was when she was half as old as the monkey will be when it is as old as its mother will be when she is four times as old as the monkey was when it was twice as old as its mother was when she was one third as old as the monkey was when it was as old as its mother was when she was three times as old as the monkey was when it was one fourth as old as it is now.

How long is the banana?

[quote="FantasyLover"]...izzy, lollerskates wants us to solve it and have fun with it...[/quote]

You guys are vicious!

I don't get it. What's the point of this?

wow....

I didn't get the question though....

Very confusing....]

 :wacko:  :wacko:

 :surrender:

That was a fairly tricky question, but after much thought, banging my head on the wall, and measuring a banana from my kitchen i came up with 5 3/4 in.  At least i hope thats right... :maybe:

I don't think I'd have fun with this ...
 :(  :(  :(

oooooh! I know! it's zero inches! 'cause the monkey ate it!  :lol:

...right?

[quote="shaun ong"]hmm.......i went and ask my father......my father went and ask my grand father.....and my grandfather went to ask my great grand father in his dreams..........at last, from my ancestors,they told methe answer is 3.75 inchest. ITs easy just measure the average length of 2 monkeys middle finger, then divide by number of total eyes then multiply by the length of the average length of a banana then power it up to  the number of words i just type[/quote]
is there a 3 eyed monkey?


integration
inequalities
LaTeX
IMC
college contests

[quote="Moubinool"]Inequality "accroissement finis"
$|g(1)-g(0)|\leq ||f'||(h(1)-h(0))$
[/quote]
Finite increment?

the problem is: you must leave a space before ":" and one after; dunno why; in the texnic program your formula would have been good, but here you must do this

[quote="LevonNurbekian"]Here is my solution:
$|\int_{0}^{1}{f(x)(f^{2}(x)-f^{2}(0))dx}|=|\int_{0}^{1}{f(x)dx{\int_{0}^{x}{2f(y)f'(y)dy}}}|=2|\int_{M}{f(x)f(y)f'(y)dxdy}|$,where $M$ is the triangle $\{(x,y) : x{\in}[0,1],y{\in}[0,x]\}$.So
$||{\leq}2{\max}{|f'|}|\int_{M}{f(x)f(y)dxdy}|={\max}{|f'|}\int_{[0,1]^{2}}{f(x)f(y)dxdy}={\max}{|f'|}(\int_{0}^{1}{f(x)dx})^{2}$.

Guys there is some problem with LaTex. :?[/quote]


Find all  cotinuous  funqtion f:R-->R such that f(f(x))=f(x)+2x for all x->R. :oops:  :oops:

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=continuous+injective&t=15554

Pierre.


A ball rolls to the right without slipping and gradually slows down to a standstill. As the ball rolls, (kinetic) friction is to the left and decelerates the velocity of the center of mass of the ball. But shouldn't it serve as a torque accelerating the angular velocity? Then how can the equation $ v_{cm} \equal{} wR$ be satisfied (and why does the ball stop rotating)?

[quote="t0rajir0u"]It is convenient to think of a ball rolling without slipping as (instantaneously) rotating about its point of contact with the floor rather than about its center of mass.  Kinetic friction contributes no torque about this point.[/quote]

If there is no torque from friction then why does the ball slow down at all?

[quote="rituraj007"]we r considering the instantaneous pt. of contact as referance pt.--the point where friction also acts,so the torque or friction is zero.
thats it :)[/quote]

I don't think that answers my (later) question...

ofc friction si acting and slowing the motion
but actually we have assumed it 2 b 0 in our frame of ref

If the ball rolls without slipping, and no resultant force or other torques , there shouldn't be translational friction force and it shouldn't  gradually slows down to a standstill.

Our assumption that friction does not act on a rolling body is ideal and cannot be realised in practice. In practice, when a body rolls, it gets deformed or compressed at the point of contact and its centre of mass shifts so that the normal reaction from the ground is not passing through the centre of mass and contributes to the opposing torque.

As we don't have any data of deformation,etc. seems to be the ideal case...


Amy's horse van holds 6 horses. Using her van, what is the minimum number of trips Amy will have to make to move 25 horses?

Level: CD Math counts

No. Its 5 :D

[hide]You just have to find $\lceil\frac{25}{6}\rceil=5$.[/hide]

sorry i am so out of it.....that makes complete sense!!! :(

[hide]Asume that $x$ is the numbers of trips Amy has to do.

$x = 25 / 6$
$x \approx 4.16$

But since Amy can't make $4.16$ trips we have to round it up to $5$.[/hide]

You don't "round it up", because there is one more horse left over she has to make one more trip to get the horse back.  It's NOT "rounding it up."


What is the constant term of the expansion of $ \left(6x\plus{}\dfrac{1}{3x}\right)^6$?

[hide=easy]Note that the x terms have to cancel, which occurs when the coefficient is 6c3. We get 6c3*6^3*x^3*1/3^3*1/x^3=160.[/hide]


$ a,b,c$ are positive real numbers then
\[ \left(\sqrt[3]{a\frac{a\plus{}b}{2}\frac{a\plus{}b\plus{}c}{3}}\right)^2\cdot \sqrt[3]{a\frac{2}{\frac{1}{a}\plus{}\frac{1}{b}}\frac{3}{\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}}}\geq \left(\frac{a\plus{}\sqrt{ab}\plus{}\sqrt[3]{abc}}{3}\right)^3\]

[quote="Sunjee"]$ a,b,c$ are positive real numbers then
\[ \left(\sqrt [3]{a\frac {a \plus{} b}{2}\frac {a \plus{} b \plus{} c}{3}}\right)^2\cdot \sqrt [3]{a\frac {2}{\frac {1}{a} \plus{} \frac {1}{b}}\frac {3}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}}}\geq \left(\frac {a \plus{} \sqrt {ab} \plus{} \sqrt [3]{abc}}{3}\right)^3\]
[/quote]

try

$ a \equal{} 1,b \equal{} 1,c \equal{} \sqrt[6]{6} \equal{} 1.3480061545972776674...$

[quote="Sunjee"]$ a,b,c$ are positive real numbers then
\[ \left(\sqrt [3]{a\frac {a \plus{} b}{2}\frac {a \plus{} b \plus{} c}{3}}\right)^2\cdot \sqrt [3]{a\frac {2}{\frac {1}{a} \plus{} \frac {1}{b}}\frac {3}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}}}\geq \left(\frac {a \plus{} \sqrt {ab} \plus{} \sqrt [3]{abc}}{3}\right)^3\]
[/quote]
We cannot compare both sides.

For $ a\equal{}b\equal{}1$ and $ c\equal{}0.5,$ we have $ LHS>RHS.$

For $ a\equal{}b\equal{}1$ and $ c\equal{}2,$ we have $ LHS<RHS.$


Apply the angle bissector law and stwart teorem in triangles ABE and ADC we find x=45

As castigioni so I could even realize especificarmelo thanks in advance.

Greetings  :D


Circle $ \Gamma_{1},$ with radius $ r,$ is internally tangent to circle $ \Gamma_{2}$ at $ S.$ Chord $ AB$ of $ \Gamma_{2}$ is tangent to $ \Gamma_{1}$ at $ C.$ Let $ M$ be the midpoint of arc $ AB$ (not containing $ S$), and let $ N$ be the foot of the perpendicular from $ M$ to line $ AB.$ Prove that $ AC\cdot CB\equal{}2r\cdot MN.$

Some hints

SC is the bisector of <ASB

N is midpoint of AB

Use the tangent of half-angle

let R be big radius, r be small. MN=x
CN=sqrt(2(R-r)x+x^2)
NB=sqrt(2Rx+x^2)
then ACxCB=2rx

personally, i think this is too easy for CGMO.  :D

according to the head of the question committee this year, who also happens to be China's IMO team leader this year.

According to him, the first 2 questions of each day (namely problems 1, 2, 5, and 6) are "warm-up" problems while the last two are meant to differentiate the good ones from the really good ones:)

You can also solve this by noting that C, S, and M are collinear.  Since they are collinear, then by power of a point, we have to prove that SC*CM=2r*MN, rearrange this to get SC/2r=MN/CM.  By similar triangles in a way, we can get the result.

My solution:
Let $O_1,O_2$ be the centers of this circle, respectively,and $O_1O_2\cap\Gamma_{1}=Q.$ We know $O_2,O_1,S$ collinear.Let $M'=SC\cap \Gamma_{2}$ and $N=M'O_2\cap AB.$ We know $\angle QCS=90,$ and $$\angle O_2M'S=\angle O_2SM'=^{\text{from tangent line theorem}} \angle QCN=90-\angle NCM'\to \angle NM'C+\angle NCM'=90,$$ then $M'\equiv M.$ Or $S,C,M$ are collinear.Alao from $PoP$ $AC\cdot CB=SC\cdot CM,$ and $$\triangle NMC\sim\triangle CSQ\to QS\cdot MN=MC\cdot SC.$$ As desired.


let 5 real numbers such that 3 of them determine a triangle.
prove that from all triangles determined  it  existe  one  at least has  acute angle .

i think this problem is  hard for you  :rotfl:

Don't expect answers so fast.

And do you mean positive real numbers?

Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 . But unlock because this topic has reply.


function
probability
probability and stats

Hi,

I'm trying to solve this and am stuck where to start...

If Z is a normal N(0,1), then process X =sqrt t * Z is continous and marginally distributed as a Normal N(0,t).

Is X a Brownian Motion and why??

Thanks[/img]

Brownian motion is a process with independent normally distributed increments: $B_{t_{2}}-B_{t_{1}}=N(0,t_{2}-t_{1})$. Your process does not have independent increments.

Sorry, I don't understand...

Agree that variance of a Brownian is equal to the length of the process.  However I don;t see why the increments of the process X are not independent.  Afterall Z is a normal N(0,1)

Recall that random variable $Z$ is simply a measurable function defined on a probability space $\Omega$. The process $X_{t}=\sqrt{t}Z(\omega)$ has increments $X_{t_{2}}-X_{t_{1}}=(\sqrt{t_{2}}-\sqrt{t_{1}})Z(\omega)$, which are very far from being independent - they are multiples of one another. (The increments $B_{t_{2}}-B_{t_{1}}$ and $B_{t_{4}}-B_{t_{3}}$ of a Brownian motion are independent when the intervals $(t_{1},t_{2})$ and $(t_{3},t_{4})$ do not intersect.)

Here's a hopefully simpler explanation: what are the sample paths of $X$? They are straight lines $t\mapsto \sqrt{t}Z(\omega)$, nothing like [url=http://classes.yale.edu/Fractals/RandFrac/Brownian/Brownian.html]Brownian sample paths[/url]. The process $X$ starts at random direction but never turns around.

How can I compute the Variance of this process X ?


geometry
3D geometry
ratio
similar triangles

[i]A closed conical contained of height 12 cm is partially filled with liquid and the level of the liquid is marked when the vertex is pointed down. When the container is inverted, it is noticed that the liquid reaches exactly the same mark. What is the depth of the liquid in the first position.[/i]

I've inferred that the liquid and empty space just switch positions and thus 1/2 the volume of the container is the volume of the liquid. The place I'm getting this discusses using similar figures and ratios to solve this problem, but I'm not sure how to set up a ratio for the height.

[quote="Andreas"][hide]We have that the two cones are similar. Then it is $\frac{r}{12} = \frac{r_{1}}{x}$ and $r_{1} = \frac{rx}{12}$
The volume of the greater cone is $V = 4\pi r^2$ and the other volume is $V_{1} = \frac{\pi r^2x^3}{432}$.
Now we know that this volume is half of $V$.
So it is $\frac{\pi r^2x^3}{432} = \frac{4\pi r^2}{2}$ hence $x^3 = \frac{12^3}{2}$ and $x = 6\sqrt[3]{4}$.[/hide][/quote]

I don't see where you got $V_{1} = \frac{\pi r^2x^3}{432}$ from.

Actually, I don't see where any of the cubes are coming from.

The volume of the cone is $V = \frac{\pi r^2h}{3}$.
So $V_{1} = \frac{\pi r_{1}^2x}{3} = \frac{\pi r^2x^2x}{12^2\cdot3}$ where we have put $r_{1} = \frac{rx}{12}$.
I hope I didn't do any mistakes.

[quote="Mosis"]
Actually, I don't see where any of the cubes are coming from.[/quote]

I think you mean cones. Anyway, the conical is the first cone, and the cone formed by the fluid in the first position is the second


MATHCOUNTS
ARML
probability
modular arithmetic
geometry
quadratics
rectangle

Hm I continued my streak of failing chapter with a 43 (27/16). Missed sprints 23, 24, and 26, all because of reading incorrectly. Yay.

Heh, my best chapter result probably was in 6th grade :P . I should probably try to get a higher score at state.

43. fail. 29 on sprint, thought #8 on target read the fraction was greater then 4. then got disqualified because I was too strong. Guess I'll have to do state next year.

i got a 42...should have gotten a 44 incorrectly read the target with the dice...and messed up #30 and some other problem on the sprint. Still made cd and states as individual tho... :P

EDIT: i failed states...its my last shot next year

45. Fail. Loss to this random guy who got 46. I skipped cd to go to arml practice

Lol, scores at chapter:

1st. 46
2nd. 45 (me)
3rd. 39 (lol.)

I got 25 on the sprint...but then I messed up so badly on the target by missing QUESTION NUMBER 1 AND 4 AND 7.  Sadly, i got a 35.  stupid target round.  messed me up in states too.  argh..

ON teh bright side, we got all 10 team questions.

Overall Difficulty:

Sprint: Average, maybe a bit easier.
Target: Everything was ridiculously easy...
Team: Same as target

Meh. 45. Missed #14 on the sprint round.

The test was a joke.

But sadly, everybody but one person from my school misread number 28 in the same way (to get an answer of $ 144 \minus{} 6$ instead of $ 144 \plus{} 6$). 45. So there was a three way tie for first (Mewto55555, tinytim, and I). :P The coordinators didn't know about tiebreakers, either, but I guess nobody cares anyway.

The sprint was pretty easy(probably easier than last year's.) #1 - 6 on the target was basically calculator brute force, and #7 - 8 was algebra bashing. I found the team really easy too. We got a 10/10 on it, in addition to drawing and coloring a pretty border of flowers around the booklet. Yay!

My thoughts:

Sprint- Definitely easier than last year
Target- 1-6 and 8 were easier than last year, but I thought 7 was harder than any target from last year.
Team- Definitely easier than last year. We got a perfect easily and our team is bad.

45  :lol: !
Missed #8 or #9 on sprint :mad: .

[quote="math154"][quote="xpmath"]Team- Definitely easier than last year. We got a perfect easily and our team is bad.[/quote]

You didn't even need your team. :wink:[/quote]

almost... On #9 I did it in an incorrect way and got that there were no solutions  :wallbash_red:

I almost started checking $ \bmod9$ on that one. :P Turns out it only eliminates $ 1$ of $ 3$ cases anyway.

[quote="MCNational!"]UGH I PHAIL

i got a 28 on sprint, 16 on target, perfect on team...
and i still got 5th written and u phailed and got 6th after cd

and my team's scores were 40,41,40,44(me :D )

we got 3rd team... :mad:[/quote]

Wow, your state must be very competitive...... :|

Lol... The first time Michigan and competitive have been used in a sentence.

My team had 41, 41, 38, 38 and we got 5th...

My team got 3rd team and

a 61.25
 :mad:

[quote="echen"]31.5 as in the average for individual or overall official team score. That means you either got like a 1 on the team or your team got like a 12 on written.[/quote]

i meant the overall team score
because i was the only person :P

I got 39 fail.fail.fail :(
25 sprint 14 target
first team

our team scores were something like this:

45 (AIME15), 43,42,39 (me)

Bleh our scores are like:

46 (me) ,43,43,43

perfect team round

so score=63.75

That was evidently first in our fail chapter....  :ewpu:

Hopefully next year's chapter test will be much harder... I remember last year, the top score was like a 58 and only one team had a perfect team round...

erm last year, our team had a score of 62.25.....

(44,44,44,37)/10

my team's scores: 45(me), 34, 28, 26
team roudn: (rest of members sang), 10

my school is extremely fail...i don't know why.
at least i'm going to nats!

[quote="LilyBreeze"]I got a 45 on the chapter by missing number 5 =)
my team got 1st, scores were 45,44, 44,43 i think, 10 on team round, our school was pretty good this year, compared to the past years.[/quote]

whoa, pwnage scores much?

[quote="randomtext"]i got a 38... (fail)

22 on sprint, it was pretty easy but i have this thing with stupid mistakes and not reading the question...

16 on target... IT WAS EASY O__O

and team was really easy also but our team realized we made a stupid mistake and changed it at the last minute  :blush:[/quote]

Wow! Nice! I got 25 in sprint and 14 in target. I really messed upon the target since I missed 7 accidentally...

[quote="jonathanchou711"]
Wow! Nice! I got 25 in sprint and 14 in target. I really messed upon the target since I missed 7 accidentally...[/quote]

That's exactly what I got (25 sprint/14 target) and that is also the same question i missed on the target (#7)

YOU CHEATER!!!

jkjk
 :D


Show that if $ a_i\geq 1$ , for $ i\in\mathbb{N}^*$ , then:

$ (1 \plus{} a_1)(1 \plus{} a_2)(1 \plus{} a_3)\cdot\ldots\cdot(1 \plus{} a_n)\geq\frac {2^n}{n \plus{} 1}\cdot(1 \plus{} a_1 \plus{} \cdot\ldots\cdot \plus{} a_n)$

[hide]It's pretty straightforward that

$ (1 \plus{} a_1)(1 \plus{} a_2)(1 \plus{} a_3)\cdot\ldots\cdot(1 \plus{} a_n)\geq2^n\sqrt {a_1a_2a_3\ldots a_n}$

But the

$ \sqrt {a_1a_2a_3\ldots a_n}\geq \frac {1 \plus{} a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n}{n \plus{} 1}$

Is what I am having trouble with. I know it is probably a real basic dumb inequality that I am missing but I really can't figure it out at the moment (probably the sleep)

[/hide]

You're probably having trouble with that inequality because the sign should be the other way. Edit: Or not... I think I'm wrong. :(

Well it is obvious it would be if LHS was $ \sqrt [n]{a_1a_2\ldots a_n}$ , but how do I show it for square root?

EDIT: Since $ a_i\geq 1$ then $ \frac{1\plus{}a_1\plus{}a_2\plus{}\ldots\plus{}a_n}{n\plus{}1}\geq \sqrt{a_1a_2\ldots a_n}\geq \sqrt[n]{a_1a_2\ldots a_n}$
Yes, it's the sleep. Thanks anyways man you spared me many minutes of pointless thinking.

unfortunately, neither inequality chain you wrote is true: for example, try $ n \equal{} 1, a_1 \equal{} 2$ and $ n \equal{} 2, a_1 \equal{} a_2 \equal{} 2$ for $ \sqrt {a_1a_2a_3\ldots a_n}\ \text{versus}\ \frac {1 \plus{} a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n}{n \plus{} 1}$.

As a hint, try substituting $ b_i \equal{} a_i \minus{} 1$ to obviate that condition.

recently I have had problems with this one, yet again. Anyone help?

You can just expand and keep in mind that $ a_i \ge 1$, or to make it nicer and more obvious, do [b]azjps[/b]'s substitution and then expand. There's probably a very awesome solution which doesn't require expanding though (this is really weak though so it doesn't matter too much).

Just use induction, let $ M_n \equal{} \frac{2^n}{n\plus{}1} \left(1\plus{}\sum_{i\equal{}1}^n a_i \right)$

$ M_{n\plus{}1} \equal{} \frac{2(n\plus{}1)}{n\plus{}2}M_n \plus{} \frac{2^{n\plus{}1}}{n\plus{}2} a_{n\plus{}1} \qquad(1)$

All you have to show is

$ (1\plus{}a_{n\plus{}1})M_n \ge M_{n\plus{}1}$

Using $ (1)$ and the fact that $ M_n \ge 2^n$ the result follows easily


percent
ratio
greatest common divisor
algebra
function
domain
Ring Theory

Note: Please keep all problems between 6th grade basic math and 8th grade geometry. Include all required formulas to solve the problems. We don't want this marathon to get locked like the other one!!! Also, no tangent, sine, or cosine ratios please. Thanks much.

[size=200]Problem 1[/size]

What is the area of a regular octagon with an apothem of 13 and a perimeter of 80?

Formula: Area of a regular polygon= 1/2 of the apothem x the height

[quote="i_like_pie"]1. A regular octagon with perimeter $ 80$ has an apothem of length $ 5 \plus{} 5\sqrt {2}$, not $ 13$.

2. In general, polygons do not have heights.

3. $ \text{Area} \equal{} as$, where $ a$ is the apothem and $ s$ is the semiperimeter.[/quote]

sorry i meant to say perimeter not height. Also i got this problem out of my geometry book. You take half the apothem and multiply it into the perimeter

Lets just have something more challenging.
Find the value of y that makes this equation have no solutions
\[ \frac{52xy\plus{}32}{21y}\equal{}12x\plus{}61\]

hmm, dude, you need to start looking at which forum you're posting in..

anyways, $ y\equal{}0$, anything else will produce some solution $ (x,y)$

let's bring down the level, even though I find that hard to do

Simplify the following expression: $ \sqrt[4]{5184}$ (what?  we did this in our math class, just I changed it from 144 to 5184)

no we didn't, anyways,
[hide]5184=2*2*2*2*2*2*3*3*3*3
to the foruth power2*2*2*2*2*2*3*3*3*3=6*(4 to the fourth power)
4 to the fourth power equals the sqt 2. So the answer is 6sqt2[/hide]

Sorry I didn't know the commands for these symbols.

Problem:How many minutes will it take a car to travel 50 miles if it always drives at 30 miles per hour.

eh, it was on our hw that I did in my head in like 2 minutes...

anyways, what is the shortest distance from the points $ (2,3)$ and $ (10,18)$?

and if you want to try, what is the shortest distance from the points $ (1,2,3,4)$ and $ (4,6,8,12\sqrt{2})$? (obviously you'll have to use the distance formula if you want to try, but you only have to answer the first one)

[quote="ZhangPeijin"]Uhh..
4 coordinates?
Is that even possible?[/quote]

4 coordinates=4 dimensions

hey, I didn't say you had to do that one :wink:

Here's a problem:

Two eight sided die are rolled. What is the probability that on both, dice the top number is have a factor of 12.

[hide="Alan first problem"]
$ \sqrt {8^{2} \plus{} 15^2} \equal{} \boxed{17}$
[/hide]

[hide="Alan second problem"]
$ \sqrt {3^2 \plus{} 4^2 \plus{} 5^2 \plus{} (12\sqrt {2} \minus{} 4)^2} \equal{} \sqrt { \minus{} 6(16\sqrt {2} \minus{} 59)}$
[/hide]

[hide="shentang problem"]
The factors of 12 is $ 1,2,3,4,6,12$ obviously you can not roll a 12 on a die so the answer is $ \frac {5}{64}$
[/hide]

Edit: my non-reading abilities strike again!

Note that on my problem it is an eight sided dice, not six, so you have the numerator correct.

Zhang Peijin's problems:

[hide="Problem 1"]

$ d\equal{}\sqrt{(10\minus{}2)^2\plus{}(18\minus{}3)^2}$

$ d\equal{}\sqrt{8^2\plus{}15^2}$

$ d\equal{}\sqrt{64\plus{}225}$

$ d\equal{}\sqrt{289}$

$ d\equal{}17$[/hide]

[hide="Problem 2"]

$ d\equal{}\sqrt{(12\sqrt{2}\minus{}4)^2\plus{}(8\minus{}3)^2\plus{}(6\minus{}2)^2\plus{}(4\minus{}1)^2}$

$ d\equal{}\sqrt{(12\sqrt{2}\minus{}4)^2\plus{}5^2\plus{}4^2\plus{}3^2}$

$ d\equal{}\sqrt{288\minus{}2(12\sqrt{2})(4)\plus{}16\plus{}25\plus{}16\plus{}9}$

$ d\equal{}\sqrt{354\minus{}96\sqrt{2}}$

It won't simplify any further than that.

For those who thought this problem was complicated, it's easy. Just add another term for every extra dimension.[/hide]

fishythefish. You can factor a bit more.

[quote="shentang"]@lawrence wu
[hide]
I just divided by random numbers that were under 36 and somehow got an answer. Dimensions are 12, 16 and 24.
[/hide]

New Problem:

two letters are chosen at random, what is the probability that it is a word. (Might need a dictionary)[/quote]

From my amazing knowledge of Scrabble, I can instantly say that there are 101 two-letter words. Out of a total 676, the probability is then $ \boxed{\frac {101}{676}}$.

[quote="dreambold22"]The surface area of a cube is 96 feet. how many cubic feet r there in the volume of the cube?[/quote]

[hide="Solution to dreambold22's question as stated above"]
$ (\sqrt {\frac {96}{6}})^3 \equal{} 64$

Thinking process:
Any (the) cube has six sides.
Area of one side of the cube is $ 96/6$.
Length of the cube?
It is the square root of the area of one side of the cube
Volume?
The cube of the length of the cube.
[/hide]

[hide="New problem"]
Determine the second smallest prime factor of
$ \frac {1^3 \plus{} 1}{1 \plus{} 1} \plus{} \frac {2^3 \plus{} 1}{2 \plus{} 1} \plus{} \frac {3^3 \plus{} 1}{3 \plus{} 1} \plus{} ... \plus{} \frac {2004^3 \plus{} 1}{1 \plus{} 1} \plus{} \frac {2005^3 \plus{} 1}{1 \plus{} 2005}$

SMO 2005, #35

This is grade 8 math

Edit: the mistake in the expression has been rectified. sorry
[/hide]

[quote="7h3.D3m0n.117"][quote="ThetaPi"]This is grade 8 math[/quote]

That most certainly is not.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=151708#151708[/url][/quote]

I'm allowed to take it and I'm in grade 7, so why isn't this out of grade 8 math, even though it's olympiad style? [No prerequisites for this competition.]

[quote="ThetaPi"]I'm allowed to take it and I'm in grade 7, so why isn't this out of grade 8 math, even though it's olympiad style?[/quote]
I'm allowed to take the IMO, but does that mean it's my level?

[quote="ThetaPi"]Determine the second smallest prime factor of
$ \frac {1^3 \plus{} 1}{1 \plus{} 1} \plus{} \frac {2^3 \plus{} 1}{2 \plus{} 1} \plus{} \frac {3^3 \plus{} 1}{3 \plus{} 1} \plus{} ... \plus{} \frac {2004^3 \plus{} 1}{1 \plus{} 1} \plus{} \frac {2005^3 \plus{} 1}{1 \plus{} 2005}$

SMO 2005, #35[/quote]
[hide="Solution"]Using $ (a \plus{} b)^3 \equal{} (a \plus{} b)\left(a^2 \minus{} ab \plus{} b^2\right)$, we can simplify the expression:
\[ S \equal{} \left(1^2 \minus{} 1\cdot1 \plus{} 1^2\right) \plus{} \left(2^2 \minus{} 1\cdot2 \plus{} 1^2\right) \plus{} \ldots \plus{} \left(2005^2 \minus{} 1\cdot2005 \plus{} 1^2\right)
\]
Grouping like terms, we get $ S \equal{} \left(\sum_{k \equal{} 1}{2005}k^2\right) \minus{} \left(\sum_{k \equal{} 1}{2005}k\right) \plus{} 2005$.

Evaluating the sums, $ S \equal{} \frac {2005\cdot2006\cdot4011}{6} \minus{} \frac {2005\cdot2006}{2} \plus{} 2005 \equal{} 2686718045$.

The smallest prime factors of $ 2686718045$ are $ 5$ and $ 11$, so the answer is $ \boxed{11}$.[/hide]

[b]Problem ThetaPi-2[/b]

[hide]
Combinatorics

Consider the linear equation
$ x_{1} \plus{} x_{2} \plus{} ... \plus{} x_{n} \equal{} r$
where $ r$ is a nonnegative integer.

Show that the number of nonnegative integer solutions to the equation is given by
$ {r \plus{} n \minus{} 1 \choose r}$ and the number of positive integer solutions to the equation is given by $ {r \minus{} 1 \choose r \minus{} n}$.

THIS IS NOT A MATH OLYMPIAD PROBLEM.

Suppose you are a 6th-grader who has no experience in math olympaids before, but has fully mastered basic algebra [no quadratics]. If you are given the definitions of binomial coefficients and the factorial, how is it possible to prove these statements?

[b]Challenge: Try to make your proof as simple as possible.[/b]
[/hide]

[quote="Lawrence Wu"][hide="Solution"]$ \left(\sqrt {\dfrac{96}{6}}\right)^3 \equal{} 64$[/hide]
A terrarium in the shape of a rectangular prism has a volume of 4608 cubic inches. Its length is more than 10 inches. Its width is w inches, its height is w+4 inches, and its length is 36-w inches. What are the dimensions of the terrarium?[/quote]
Solve it mathematically (It was solved by guess-and-check earlier).
Hint: Factor by grouping. It's in my school Algebra I textbook.

[quote="fishythefish"]I'm pretty sure that although you're trying to make this look like a 6th grader could solve it, this is getting out of classroom math range, and will get locked like the other marathon. Just because a gifted 6th grader could potentially solve a problem, it doesn't make this classroom math.[/quote]
I think that a better solution would be to simply delete the posts that are out of classroom math range.
@ThetaPi: If you don't learn anything about it in grade 8 or less, in the normal curriculam, then you shouldn't post anything about it.

Heres a knew question:
23. Catrina runs 100 m in 10 seconds. Sedra runs 400 m in 44 seconds. Maintaining these constant
speeds, they participate in a 1 km race. How far ahead, to the nearest metre, is the winner as she
crosses the finish line?
(A) 100 m (B) 110 m (C) 95 m (D) 90 m (E) 91 m

Gauss contest for grade 7 2001

[hide]
Catrina runs 1km in 100 seconds. Sedra runs 1km in 110 seconds. So Catrina wins the race by 10 seconds. Thus she goes 100 meters.
[/hide]

When you solve a problem, post a new one. When you post a new problem, solve the previous problem.
It seems my problem was skipped again.

Ok, I will post new problem.

[hide="extremely hard (well not really - but not classroom math level) problem - someone post another please"]
You have two blocks of clay in cube form and the edges are 10 cm. How many hollow spheres with an interior radius of 5 cm can you make with that amount of clay, if you must make the thickness of the clay ½ (cm.)?
[/hide]


No!!! The answer is zero.  If you don't believe me, cancel out the fractions and then multiply it on a calculator.  I keep getting 1!! Ahhh!!

EDIT: This wasn't a double post originally, Joe deleted his faulty answer of 1...

[quote="mysmartmouth"]$\frac{30!}{20!\cdot10!}$
How many zeroes does this number end in?[/quote]

Well, $\frac{30!}{20!\times 10!}$ leaves $\frac{30\times 29\times 28\times...\times 21}{10\times 9\times 8\times...\times 1}$

I got 0, and I think it has something to do with the number of 2's on either side of the fraction bar...they both have 8 (the same number) :o

Since there are no 2's, there are no 0's.  I can't really explain this though :( I guess it's just a very [i]special[/i] problem that isn't as easy as it looks. :)

Yes.  30! has 26 twos, as does 10!20!.  Therefore, they all cancel.  Nice problem!

[quote="rs9999"][quote]Yes, you have to find the number of factors of 5, then the number of factors of 2, and finally take the minimum.[/quote]you dont need to find the number of 2s. There will always be more of them than 5s.[/quote]

no, not if they cancel out more twos than fives ;) , which happens in this case


Putnam
inequalities proposed
inequalities

how should the number $1,2,...,n$ be arranged in a cicle to make the sum of the products of pairs of adjacent numbers as large as possible? as small as possible?
thank you very much.

The first part is Putnam 1996 B3 problem. The second part can be done in the same way. The solution I've also found and I do not want to write it here you can find on kalva.