Challenging Math Problems (Or not)

[winneratlife]

Ok from now on if i post a geometry question, ill post a picture with it, ill have the pic up in a few minutes.

Guys how do i paste a pic to a post directly? I hate having to link you guys to photobucket...

http://s658.photobucket.com/albums/uu30 ... oblem1.png

There you go, the smaller circle is O and the bigger is Q, find the ratio of O to Q

The ratio of the radii is one to nine.

Meh, this looks fun!

So, find:



Because this is bashable, you must explain how you did it. NO brute force allowed!

[lllazar]

Ok from now on if i post a geometry question, ill post a picture with it, ill have the pic up in a few minutes.

Guys how do i paste a pic to a post directly? I hate having to link you guys to photobucket...

http://s658.photobucket.com/albums/uu30 ... oblem1.png

There you go, the smaller circle is O and the bigger is Q, find the ratio of O to Q

The ratio of the radii is one to nine.

Meh, this looks fun!

So, find:



Because this is bashable, you must explain how you did it. NO brute force allowed!

Uhh, your on the right track with my question, but i asked you to find the ratio of the areas...ill give it to you though, its 1:81, you just had to square the radii to find the ratio of the areas.

[lllazar]

Ok from now on if i post a geometry question, ill post a picture with it, ill have the pic up in a few minutes.

Guys how do i paste a pic to a post directly? I hate having to link you guys to photobucket...

http://s658.photobucket.com/albums/uu30 ... oblem1.png

There you go, the smaller circle is O and the bigger is Q, find the ratio of O to Q

The ratio of the radii is one to nine.

Meh, this looks fun!

So, find:



Because this is bashable, you must explain how you did it. NO brute force allowed!

I think i got it....you factor each pair like this:
becomes

As i did this for each "pair", i found that the sequence decreases by 4 each time like this:

(98-97)(98+97) = 195
(96+95)(96-95) = 191

So then i went to the 51st number in the sequence, like this:

When i did 199-(51*4), i got -1. So i knew the sequence ended with 3, since -1 + 4 = 3.

The answer is 199+195+191+187...+3, or 5050.

[carneyf1d]

with partial sums of the series equation
Summation from 0 to 100 of (-1)^n * n^2 = 1/2 *(-1)^n * n *(n+1)
= 5050

[lllazar]

with partial sums of the series equation
Summation from 0 to 100 of (-1)^n * n^2 = 1/2 *(-1)^n * n *(n+1)
= 5050

Wow that looks like a much simpler way to do it

Could you explain it to me?

[carneyf1d]

yea basically you can write the sequence as a series from 0 to n. In this case n = 100. the series is is equal to summation (-1)^n * n^2 from 0 to 100.
you can split this up into two series being multiplied together. So it becomes = summation (-1)^n from 0 to 100 times the summation n^2 from 0 to 100.
summation of (-1)^n = 1/2 *(-1)^n
summation of n^2= n*(n+1)
when you plug in n = 100 and multiply the two summations, you get 5050.

[lllazar]

yea basically you can write the sequence as a series from 0 to n. In this case n = 100. the series is is equal to summation (-1)^n * n^2 from 0 to 100.
you can split this up into two series being multiplied together. So it becomes = summation (-1)^n from 0 to 100 times the summation n^2 from 0 to 100.
summation of (-1)^n = 1/2 *(-1)^n
summation of n^2= n*(n+1)
when you plug in n = 100 and multiply the two summations, you get 5050.

Ahh, thanks!

[winneratlife]

I think i got it....you factor each pair like this:
becomes

As i did this for each "pair", i found that the sequence decreases by 4 each time like this:

(98-97)(98+97) = 195
(96+95)(96-95) = 191

So then i went to the 51st number in the sequence, like this:

When i did 199-(51*4), i got -1. So i knew the sequence ended with 3, since -1 + 4 = 3.

The answer is 199+195+191+187...+3, or 5050.

llazar, your way is actually much easier...the only thing is...

If you do all of the difference are 1s and can be eliminated, leaving which is clearly 5050.

Also, carney, I don't get what you did. I believe the sum of is actually is it not?

EDIT: carney, you gonna post a problem? Otherwise, I'll be mean and make you guys do mass points! (hehe)

[binary010101]

Find the formula for the volume of the right-angle intersection of two right circular cylinders without using calculus.

http://en.wikipedia.org/wiki/File:Bicyl ... _solid.gif

EDIT: It was in my AP Calc textbook, and I still can't figure it out...

[winneratlife]

Find the formula for the volume of the right-angle intersection of two right circular cylinders without using calculus.

http://en.wikipedia.org/wiki/File:Bicyl ... _solid.gif

EDIT: It was in my AP Calc textbook, and I still can't figure it out...

Ouch...

Give me a while to work on it if I have time...Dang...