Circles

Forums: Riddles
Email this Topic • Print this Page

Consider a circle whose center is (2,2) and whose radius is 1, and the straight line that goes through the origin and that is tangent to this circle so that the intersection between them is as shown in the picture below.With this new point we make a new circle whose radius is half of the first one, and we calculate the corresponding intersection point with the same suppositions as in the first case.We repeat the process to the infinite. Find the distance between the center of the circle in the infinite and the origin (point (0,0)).

http://img2.freeimagehosting.net/uploads/th.2d26424a6e.jpg

Topic Stats
Top Replies
Link to this Topic

Type: Discussion • Score: 1 • Views: 1,017 • Replies: 17

No top replies

You're absolutely pathetic, go away.

0 Replies

Re: Circles

official wrote:

Consider a circle whose center is (2,2) and whose radius is 1, and the straight line that goes through the origin and that is tangent to this circle so that the intersection between them is as shown in the picture below.With this new point we make a new circle whose radius is half of the first one, and we calculate the corresponding intersection point with the same suppositions as in the first case.We repeat the process to the infinite. Find the distance between the center of the circle in the infinite and the origin (point (0,0)).

How long a time should I devote to considering this circle? And what am I considering it for? Is it up for an award? Will it receive a gold or silver star if I complete your homework for you? Nope, ain't gonna happen.

0 Replies

It's not homework.

0 Replies

official, we get a lot of people coming in here asking us to do homework. We don't do homework. When you sign up as a random member and post 5 questions that all sound like homework questions...it raises flags. If you have a plausible explanation for why you are posting this flurry of homework-sounding-questions, you'll be helped.

0 Replies

The tangent is perpendicular to the radius at the tangent. Ignore the circles and think about right triangles where the hypotenuse of one (origin to the tangent)becomes the long leg of the next and the short leg of one is half the short leg of the previous. Throw in some Pythagoras and an infinite geometric series and you get sqrt(28/3).

0 Replies

How odd

.. your question matches this word for word... and there's the answer... (Mark found it first)

http://mathcentral.uregina.ca/QQ/database/QQ.09.05/paul5.html

Riddle me this ... did we pass

0 Replies

It seems right, but the riddle site will not accept that answer :\

0 Replies

timberbranch wrote:

How odd .. your question matches this word for word... and there's the answer... (Mark found it first)

http://mathcentral.uregina.ca/QQ/database/QQ.09.05/paul5.html

Riddle me this ... did we pass

Actually, Mark solved it on his own. It wasn't difficult.

0 Replies

markr wrote:

The tangent is perpendicular to the radius at the tangent. Ignore the circles and think about right triangles where the hypotenuse of one (origin to the tangent)becomes the long leg of the next and the short leg of one is half the short leg of the previous. Throw in some Pythagoras and an infinite geometric series and you get sqrt(28/3).

from the link :

"The circles are not really relevant -- you just need the fact that a radius is perpendicular to the tangent at its point of contact. You get a sequence of right triangles, so use the Pythagorean theorem to get what is called a geometric series"

Rolling Eyes

0 Replies

Sheesh! Of cause they are similar; they are both solving the same problem.

"Throw in some Pythagoras and an infinite geometric series "

Tell me; what else are you gonna use to come up with the answer?

When you have been around longer than five minutes you may come to appreciate the quality of Mark's answers. If on the other hand, you are making some sort of cryptic allegation - go forth and multiply.

BTW Welcome to the forum

0 Replies

Tryagain wrote:

When you have been around longer than five minutes you may come to appreciate the quality of Mark's answers.

Since my measuring stick isn't up to snuff for this comparison, I'll search for more quality in a different answer. The two looked about the same to me ... not quite STRIKING, but close enough for a newbie. Sorry Mark, good to have friends 'round here ... I guess.

0 Replies

I was told that the format sqrt (28/3) was correct, but the values were a little too large.

0 Replies

Thanks Try, but I'm not entirely worthy of the kind comments. I made a mistake.

Here's what I did:

Code:

Long Leg   Short Leg   Hypotenuse

---------------------------------

sqrt(8)        1       sqrt(8+1)

sqrt(8+1)     1/2      sqrt(8+1+1/4)

sqrt(8+1+1/4) 1/4      sqrt(8+1+1/4+1/16)

and so on

The distance is sqrt(8 + sum(1 + 1/4 + 1/16 + 1/64 + ...))
1 + 1/4 +1/16 + 1/64 = 1/(1 - 1/4) = 1/(3/4) = 4/3
8 + 4/3 = 28/3
Distance = sqrt(28/3)

The mistake was that I reversed the roles of the long leg and the hypotenuse. It should be:

Code:

Hypotenuse  Short Leg  Long Leg

-------------------------------

sqrt(8)        1       sqrt(8-1)

sqrt(8-1)     1/2      sqrt(8-1-1/4)

sqrt(8-1-1/4) 1/4      sqrt(8-1-1/4-1/16)

and so on

The distance is sqrt(8 - sum(1 + 1/4 + 1/16 + 1/64 + ...))
1 + 1/4 +1/16 + 1/64 = 1/(1 - 1/4) = 1/(3/4) = 4/3
8 - 4/3 = 20/3
Distance = sqrt(20/3)

By the way, the other site had it wrong also, but in a different way. It had:
8 - (1 - 1/4 - 1/16 - 1/64 - ...)

This equates to 8 - (1 - 1/3) = 8 - 2/3 = 22/3

0 Replies

I got that far right before you posted that second time. Thanks for all the help though, you really and I mean really pushed me in the right direction!

0 Replies

Does that mean sqrt(20/3) was accepted?

0 Replies

Yes

and thanks again!

0 Replies

Circles

Related Topics

Quick Links

My Account

able2know