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By Tongos (Tongos) on Monday, July 19, 2004 - 06:36 pm: Edit |

calling all math lovers and contestants!

Tongos Official Math Competition will be held, covering geometry, algebra, arithmetic, and calculus i, ii, or iii.

purpose of the TOMC: to get more people interested in problem solving and math!

what i need is a list of people who will be registering and people who will be asking the questions. people who ask there own questions can't give the answer to that question obviously.

note: if you ask questions, be sure to make the questions thorough and understandable to the contestants. questions out of books is great, sat tests, ap tests, amc, aime, usamo, act.. ect would be great.

1st round- easiest (90% of the contestants should live)

2nd round- easy to moderate. (50% should live)

3rd round- moderate (not many should come out alive)

4th round- hard (barely anybody should come out)

5th round- hardest (winner!)

each round will have five to ten questions or so, and to give everybody a chance, questions should be asked to specific people. if they don't have an answer within a time limit, then it goes on to the next person. so once you join, its important to stick in the competition!

the length and the aliveness of the game will be based on you guys.

so sign in if you want to go into the competition,

again, if any have an idea of how the competition should be played, i would love to consider any changes. right now, i think it would be better if we discuss the competition rules first

hope this works out! even when tongos isn't around

By Jenesaispas (Jenesaispas) on Monday, July 19, 2004 - 06:45 pm: Edit |

How will each person submit his answers without seeing other posts?

The competition sound fun , even though I'm not a math expert... I just like to do it recreationally.

By Adidasty (Adidasty) on Monday, July 19, 2004 - 06:50 pm: Edit |

I'm in.

By Tongos (Tongos) on Monday, July 19, 2004 - 07:04 pm: Edit |

to answer jenesaispas question, a question will be asked to a specific person, if they dont get it right or dont answer it, it moves on to the next person. they simply write down the answer, with no work. if they get it wrong, it moves onto the next person. until somebody gets it right, 10 points. then we move on to the next question and start with the person after the person who got it right and goes in this circular fashion. i really hope that its not only me and adidasty and jenesaispas, that would kind of put a lid on the whole idea.

By Jshifton (Jshifton) on Monday, July 19, 2004 - 07:08 pm: Edit |

Sounds fun. I'm in.

By Jaug1 (Jaug1) on Monday, July 19, 2004 - 07:31 pm: Edit |

Sounds good. I'm up for it, but when is it taking place?

By Ilovefood (Ilovefood) on Monday, July 19, 2004 - 07:40 pm: Edit |

hm im terrible at math, but id like to help get your little contest rolling, so count me in.

By Phantom (Phantom) on Monday, July 19, 2004 - 08:22 pm: Edit |

I'm in. Trying to brush up on my math problem-solving skills this summer...

By Supernal_Being (Supernal_Being) on Monday, July 19, 2004 - 09:17 pm: Edit |

Hola, what's up? I'm a decent mathematician. Not amazing or anything, but I'd love to give it a try. And besides, I LOVE MATH!!! Hehe. Sorry. But yeah, count me in. Tell me when we start and all that other good stuff. And if you have to give a problem to only me, my email is not what's listed on the profile, but htown1397@hotmail.com. NICE THREAD. Good luck everyone.

By Jenesaispas (Jenesaispas) on Monday, July 19, 2004 - 09:24 pm: Edit |

I'm in then!

By Jaredthegreat (Jaredthegreat) on Monday, July 19, 2004 - 11:29 pm: Edit |

Sure, why not...

By Feuler (Feuler) on Monday, July 19, 2004 - 11:37 pm: Edit |

Count me in.

By Supernal_Being (Supernal_Being) on Tuesday, July 20, 2004 - 10:24 am: Edit |

More people must participate, or the competition must start. Hence, BUMP.

By Anjin (Anjin) on Tuesday, July 20, 2004 - 10:50 am: Edit |

I'm in.

By Conker (Conker) on Tuesday, July 20, 2004 - 11:24 am: Edit |

I think I'll pass, but I thought this was a thread discussing the national mathematics competition of Tonga when I first saw the title. Serious.

By Devious (Devious) on Tuesday, July 20, 2004 - 01:09 pm: Edit |

I second Anjin.

By Bard (Bard) on Tuesday, July 20, 2004 - 01:25 pm: Edit |

I'm in

By Adidasty (Adidasty) on Tuesday, July 20, 2004 - 01:37 pm: Edit |

when is the deadline so we can get this baby started?

By Inopa (Inopa) on Tuesday, July 20, 2004 - 01:50 pm: Edit |

in

By Piman3141 (Piman3141) on Tuesday, July 20, 2004 - 03:02 pm: Edit |

ill take a swing at it

By Tongos (Tongos) on Tuesday, July 20, 2004 - 06:32 pm: Edit |

We commence!!!. so let me get the count of whos in, weve go piman, inopa, feuler, bard, anjin, jared the great, phantom, and a bunch of others.

the first round will commence as of now, i'll ask the first question and it goes to adidasty (first to sign up)

Old SAT question. easy round.

if j and k are integers and j+k=2j+4, which of the following must be true?

I. j is even

II. k is even

III. k-j is even

no response in 5 minutes, goes to jenesapias (spelled name wrong)

By Tongos (Tongos) on Tuesday, July 20, 2004 - 06:35 pm: Edit |

just to make sure, is everybody ready. maybe i started to soon, or sudden. 3:34. i could cancel this question, for fair purposes.

By Adidasty (Adidasty) on Tuesday, July 20, 2004 - 06:36 pm: Edit |

okay so i have to answer?

By Adidasty (Adidasty) on Tuesday, July 20, 2004 - 06:39 pm: Edit |

i dunno, i am a dumbass. i got the answer to be III. lol maybe it exaplaing why i got a 690??? lmfao

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 06:39 pm: Edit |

I'm in

By Adidasty (Adidasty) on Tuesday, July 20, 2004 - 06:49 pm: Edit |

my answer is III. if that is the correct answer, tongos, please notify me. lol

By Tongos (Tongos) on Tuesday, July 20, 2004 - 06:52 pm: Edit |

alright the answer is d. you all correct all the way, ten points for adidasty.

By Tongos (Tongos) on Tuesday, July 20, 2004 - 06:52 pm: Edit |

i meant three. yeah your still correct

By Tongos (Tongos) on Tuesday, July 20, 2004 - 06:56 pm: Edit |

okay another question, with roughly the same difficulty for jenesaipas (im so sorry)

The average of 5 positive integers is 350. Two of the integers are 99 and 102 and the other integers are greater than 102. If all five integers are different, what is the greatest possible value for any of the five integers?\

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:01 pm: Edit |

okay, im tired, first person (registered) to answer it gets it.

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 07:07 pm: Edit |

I got 1342

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 07:08 pm: Edit |

I think it is 1342 (1549 - 103 - 104)

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:08 pm: Edit |

nicely done, 10 points for jared.

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:11 pm: Edit |

3rd question, with roughly the same difficulty as the last.

the distance between the points (1,2,4) and

(-3,7,5) is?

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 07:19 pm: Edit |

Is this for anyone in particular? or first to answer...? if so I got ****

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 07:19 pm: Edit |

Is this for anyone in particular? or first to answer...? if so I got ****

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:22 pm: Edit |

yes.

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 07:23 pm: Edit |

Sorry about the duplicate post.

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:23 pm: Edit |

yes, to the first to answer

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:24 pm: Edit |

anybody, jared do you have the answer. if so, dont be afraid to say it,.

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 07:27 pm: Edit |

I got SQRT( deltaX^2 + deltaY^2 + deltaZ^2)

=SQRT( 4^2 + 5^2 + 1^2)

=SQRT( 16 + 25 + 1) = SQRT(42) = 6.48?

Were you expecting an integer answer?

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 07:27 pm: Edit |

I got the square root of 42...

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:28 pm: Edit |

you are correct optimizerdad, ten points for him.

By Adidasty (Adidasty) on Tuesday, July 20, 2004 - 07:30 pm: Edit |

my answer is. 4sqrt10.5, or the sqrt.42

By Adidasty (Adidasty) on Tuesday, July 20, 2004 - 07:31 pm: Edit |

darn you guys are quick. lol

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:34 pm: Edit |

next question (still round 1 easy)

In a certain shop, items were put in a showcase and assigned prices for january. Each month after that the price was 10 percent less than the price of the previous month. if the price of an item was p dollars for january, what was the price for april. answer is in terms of p.

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 07:35 pm: Edit |

p * .90 ^ 3

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 07:36 pm: Edit |

I got .9(.9(.9(p)))=.729p

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:38 pm: Edit |

nope/

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:39 pm: Edit |

jared got it right

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 07:39 pm: Edit |

? My answer is the same as his

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:40 pm: Edit |

lengendofmax, how the heck do you change your answers.

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 07:41 pm: Edit |

after the edit...(yes I do believe you may have made a typo)

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 07:43 pm: Edit |

I had a typo, but you can have it. (I used the edit button)

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:43 pm: Edit |

weve got these numbers

Adidasty (10 points)

jared (20 points)

Optimizerdad (10 points

I would give lengendofmax his points, only if he truly earned them.

round 2 is commencing!

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 07:44 pm: Edit |

Truly earned! LOL, I just made a typo! :X

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:45 pm: Edit |

next question on the thread, a little harder, but still easy.

the sides of a triangle have the sidelengths 10,9,8 what is the area of the triangle?

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 07:47 pm: Edit |

Arargh. I seem to remember a formula like

Area = SQRT(s(s-a)(s-b)(s-c))

= SQRT( 27(17)(18)(19))

= SQRT(156978)

= 396.2

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 07:48 pm: Edit |

Nevermind

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:50 pm: Edit |

optimizerdad is correct! another 10 points

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:52 pm: Edit |

round 2, question 2.

the second term of a geometric sequence is 12 and the fourth term is 192. which of the following could be the sum of the first six terms.

a.882,b. 1023, c.4095,d. 5000, e. 6400.

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 07:54 pm: Edit |

c

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 07:55 pm: Edit |

No, darn it. Area = 0.5 (base)(height)

= (0.5)(10)(6.83) = 34.2

I figured the height h = 6.83 by dropping a perpendicular from the vertex of the triangle to the base, cutting the base into sections x and (10-x) units long, and then using Pythagorean logic.

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:55 pm: Edit |

c is correct, legendofmax, ten points.

By Bard (Bard) on Tuesday, July 20, 2004 - 07:57 pm: Edit |

The terms are 3,12,48,192,768, and 3072 so the answer is c

By Tongos (Tongos) on Tuesday, July 20, 2004 - 07:57 pm: Edit |

yeah, i'll be sure, to check my solution next time, not just look at work!

square abcd is inscribed within a sphere of radius 2. the center of the sphere, point o, is contained in the square abcd. what is the area of the square.

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 07:57 pm: Edit |

My answer is C

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 07:59 pm: Edit |

8?

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 08:00 pm: Edit |

Area of square = 2 r^2 = 8

By Tongos (Tongos) on Tuesday, July 20, 2004 - 08:00 pm: Edit |

legend of max gets the answer again!

By Bard (Bard) on Tuesday, July 20, 2004 - 08:00 pm: Edit |

too late

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 08:00 pm: Edit |

I got 2

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 08:01 pm: Edit |

I did it the lamer's way: used the radius as half a diagonal, so it makes a triangle with legs of length 2 and 2 for an area of 2, and there are four of them so 8

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 08:02 pm: Edit |

Tongos:

I'll have to leave now, duty calls. One suggestion - pl. use labels Q1, Q2 etc for the questions, so responders can label their answers as ANS(Q1) etc. Makes it easier to track.

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 08:02 pm: Edit |

oops. didn't finish the problem....

By Tongos (Tongos) on Tuesday, July 20, 2004 - 08:05 pm: Edit |

f(x)=x^2+1 and g(x)=x-5 and g(f(x)=0 then x could be

a. -3

b. -2

c. 0

d. 4

e. 6

up for grabs, and this time, unlike that triangle one, i know the answer right off the bat.

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 08:06 pm: Edit |

b?

By Bard (Bard) on Tuesday, July 20, 2004 - 08:07 pm: Edit |

b

By Tongos (Tongos) on Tuesday, July 20, 2004 - 08:10 pm: Edit |

b is the correct answer.

now onto the harder stuff. some amc 12 problems for round 3. commence yourselves.

the product of N of three positive integers is 6 times their sum, and one of the integers is the sum of the other two. find the sum of all possible values of N.

still keeping track of scores

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 08:12 pm: Edit |

"the product of N of three positive integers" what does this mean?

By Bard (Bard) on Tuesday, July 20, 2004 - 08:14 pm: Edit |

yeh i dont understand the meaning of the phrase...

By Tongos (Tongos) on Tuesday, July 20, 2004 - 08:15 pm: Edit |

the product N, sorry guys

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 08:15 pm: Edit |

336

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 08:18 pm: Edit |

How did you arrive at that answer?

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 08:18 pm: Edit |

ABC=6(A+B+C)

ABC=6*2A

BC=12

B=12/C

It has to be all whole numbers, so it's

C=1,2,3,4,6, or 12

B=12,6,4,3,2, or 1 (respectively)

A=13,8,7,7,8,13

there are three different combinations: 1*12*13=156, 2*6*8=96, 3*4*7=84

156+96+84=336

By Tongos (Tongos) on Tuesday, July 20, 2004 - 08:19 pm: Edit |

jared is right, get's another ten points.

I'm retiring for right now guys, you can continue the competition by asking questions, but for right now, im not in.

heres the scores

adidasty- 10 points

jared- 30 points

optimizerdad- 20 points

lengendofmax- 30 points

10 per answer.

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 08:22 pm: Edit |

Let N be the greatest multiple of 8, with no two identical digits. What is the remainder when you divide N by 1000? (I actually got this one when I originally saw it )

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 08:22 pm: Edit |

AB(A+B) = 6(A+B+A+B)= 6(2A+2B)=12(A+B)

so does this mean AB is 12? So all combos of 12 times 12 added together:

12 * (12 + 1)+

12 * (3 + 4)+

12 * (6 + 2)+

= 336

DARN I had this.

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 08:24 pm: Edit |

nevermind

By Bard (Bard) on Tuesday, July 20, 2004 - 08:27 pm: Edit |

4

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 08:29 pm: Edit |

120

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 08:30 pm: Edit |

I think it's better to just repost than to edit your posts. It just looks more honest.

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 08:30 pm: Edit |

ok, I will do that from now on

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 08:31 pm: Edit |

Optimizerdad, you are correct.

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 08:31 pm: Edit |

BTW it's 210

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 08:32 pm: Edit |

Are you sure? Is N 9876543210?

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 08:34 pm: Edit |

Explanation:

If a number is divisible by eight, then its last three digits are divisible by eight.

We are looking for the largest that is NOT a repeat, therefore it'll be 9876543***

There are six diferent combinations of the numbers 2,1,0, but only 120 is a multiple of 8. 9876543120/1000=9876543 R120

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 08:37 pm: Edit |

Weird my calculator is lying to me. Apparently when 9876543210 is divded by 8, it is 1234567901, but of course multiplying that by 8 should have a units digit of 8... what the.

By Joshjmgs (Joshjmgs) on Tuesday, July 20, 2004 - 08:42 pm: Edit |

Tongos, concerning this question:

the sides of a triangle have the sidelengths 10,9,8 what is the area of the triangle?

Please check over, because optimizerdad answered:

Area = SQRT(s(s-a)(s-b)(s-c))

= SQRT( 27(17)(18)(19))

= SQRT(156978)

= 396.2

In Heron's formula, s is the semiperimeter, which is in this case 13.5, not the perimeter, so the answer is NOT correct.

Sorry optimizerdad, but it should be:

SQRT((13.5)(3.5)(4.5)(5.5))

SQRT(1169)

= 34.2

This area makes a lot more sense than almost 400.

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 08:42 pm: Edit |

Legend:

1234567901 * 8 should give you a number ending in 8, not 0. Sounds like a precision problem.

Oh, and going back to that area_of_triangle problem - that formula of SQRT(s(s-a)(s-b)(s-c)) was right, but s should be (a+b+c)/2, not the (a+b+c) I had assumed. Using the correct formula,

we get s = (10+9+8)/2 = 13.5

and area = SQRT(s(s-a)(s-b)(s-c))

= SQRT( 13.5(3.5)(4.5)(5.5))

= SQRT(1169.44)

= 34.2

By Bard (Bard) on Tuesday, July 20, 2004 - 08:44 pm: Edit |

Four ants occupy the four corners of a square with side of length 1. At the same time each ant starts walking at the same speed directly toward the ant on his left. Eventually all four ants will converge at the center of the square. What path does each ant follow and what is the distance each ant walks until he reaches the center?

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 08:47 pm: Edit |

I used to be able to do quick double_checks of math answers like this, before the bulk of my gray matter had expired. For example, the area of a triangle with sides (10,9,8) < area of a (10,10,10) triangle

< area of a right_angled triangle with sides (10,10, 14.14)

< 0.5( area of a 10x10 square) or 50,

which would have been an immediate red_flag that my original answer was wrong.

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 08:50 pm: Edit |

Bard:

Time = 1/speed . Each ant is always approaching his/her neighbor at right angles to the neighbor's path, so the original distance of 1 is being reduced by a constant #units/second. Each ant walks 1 length unit before they all collide.

All the paths spiral inwards to the center of the square.

(..but to be fair, I'd seen this problem before)

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 08:52 pm: Edit |

I don't think is the same level of difficulty, but anyhow...

They all follow a spiral path.

For some reason I'm inclined to say it's the golden proportion. But since I can't exactly remember the golden proportion, I'm going with the silver, 1 + square root 2

By Joshjmgs (Joshjmgs) on Tuesday, July 20, 2004 - 08:58 pm: Edit |

1.618 or a number similiar to that is the divine proportion if thats what youre trying to remember.

I should have entered this competition, nothing too challenging so far.

By Bard (Bard) on Tuesday, July 20, 2004 - 08:59 pm: Edit |

Sorry Jared,

Try again and apply what optimizer dad said.

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 09:00 pm: Edit |

You still can...you'll just be 30 points behind a few of us...

By Bard (Bard) on Tuesday, July 20, 2004 - 09:00 pm: Edit |

Sorry Josh,

Try again

By Joshjmgs (Joshjmgs) on Tuesday, July 20, 2004 - 09:03 pm: Edit |

Oh no, I'll just do the problems on my own, no problem at all. Thats not what I was insuating at all. I can't be online much to participate anyways.

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 09:06 pm: Edit |

Josh was just telling me what the proportion I was asking for was. He was submitting that as his answer. (thanks Josh)

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 09:09 pm: Edit |

Hey, people:

C'mon - this should be less of a competition than a chance to see how other people approach problems. Let's not make 'scoring the most points' as our absolute goal here.

By Tongos (Tongos) on Tuesday, July 20, 2004 - 09:12 pm: Edit |

sorry about the triangle, i got the answer by doing my calculations, but really didnt trust it, i did it like in two seconds. and poorly scanned over optimizerdads work

the base will be 8 and the sides will be 9 and 10. okay

10^2-(8-x)^2=9^2-x^2

x will be part of the base

x=2.8125

squareroot of (9^2-(2.8125^2)) will give the height.

and you multiply it by 8 which is the base.

now divide by two to get the area of the triangle.

feel free to continue the competition, ask questions, continue! have fun!

By Tongos (Tongos) on Tuesday, July 20, 2004 - 09:24 pm: Edit |

agree, optimizerdad, agreed.

Just a question that I just thought of right now, i dont have the answer yet though, just thought of it.

the sides of a pentagon are 13,14,15,16,17 what is the area of this pentagon?

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 09:39 pm: Edit |

Tongos:

I'm not sure this pentagon is unique. Let AB=base=17, BC=16, CD=15, DE=14 and EA=13. Draw the base AB, then a circle1 with (center=A, radius=13) and circle2 with (center=B, radius=16). Seems like I could draw an arc1 centered anywhere on circle1 with arc_radius=14, and arc2 centered anywhere on circle2 with arc_radius=15, and the intersection of (arc1, arc2) would give me point D for the pentagon. (The arc_centers themselves would be points E and C).

By Supernal_Being (Supernal_Being) on Wednesday, July 21, 2004 - 03:37 am: Edit |

Oh man! You guys started, and I had no idea! Any chance I could still join??

By Tongos (Tongos) on Wednesday, July 21, 2004 - 12:17 pm: Edit |

i think we could start the competition up again today too, but i need some other people asking questions also, not just me.

By Tongos (Tongos) on Wednesday, July 21, 2004 - 03:47 pm: Edit |

bump

By Bard (Bard) on Wednesday, July 21, 2004 - 07:24 pm: Edit |

I have a very difficult problem

What is the mean distance between two random points in a unit square?

By Piman3141 (Piman3141) on Wednesday, July 21, 2004 - 11:16 pm: Edit |

radical 2 over 2? assuming each side is equal to one of course.

By Tongos (Tongos) on Thursday, July 22, 2004 - 06:06 pm: Edit |

how about a circle with radius of squareroot of two, what's the average distance between any two points?

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