Math Is Fun Forum

clooneyisagenius

Member

Registered: 2007-03-25

Posts: 56

college geometry help with proofs?

1. Prove no line can intersect a circle more than twice.
2. Assume that any straight line drawn from a point on the circle to a point inside will intersect the circle at least one more time. Prove that it will intersect the circle exactly once more.

for 1... I did it using Pasch's postulate but now the prof. wants us to do it again with a different method and I just haven't figured it. I have 3 pages of drawings and attempts but it's due tomorrow now... and the past week I've yet to figure it out???

and for 2 i feel like i need to use the fact that "no 3 lines can intercept a line from a point such that all lines have the same length" [which i've proved in this same problem set].

Ideas?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: college geometry help with proofs?

Just as a disclaimer, I haven't taken a formal course in college geometry.  Is this analytical geometry?  If so, have a general equation for the line, a general equation for the circle, then show that there can be at most 2 possible solutions.

For 2, would you consider a point on the circle to be inside it?  Otherwise, I don't see how this is possible.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: college geometry help with proofs?

I would prove that all vertical lines through a circle are symmetrical to all lines at another angle, such as horizontal, or at any angle.  If you can prove this, then you can use Ricky's vertical line solutions to prove only 0, 1, or 2 points are solutions to the circle for any x value.
I can't do a formal proof, but that is my silly idea.

---

igloo myrtilles fourmis

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: college geometry help with proofs?

You don't need vertical lines John.

(x-k)^2 + (y-h)^2 = r^2  #equation of the circle

x^2 - 2xk + k^2 + y^2 - 2yh + h^2 - r^2 = 0 #same equation expanded

ax + by - c = 0 #equation of the line

x = (-by + c) / a #solved for x

((-by + c) / a)^2 - 2((-by + c) / a)k + k^2 + y^2 + 2yh + h^2 - r^2 = 0 #plug in to circle

It is easy to see that there are at most 2 solutions to this equation when solving for y because it is a polynomial of degree 2.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: college geometry help with proofs?

Sorry Ricky, I mistakenly didn't see your "general" equation of a line, and thought
you intended to prove that for a circle, for each x, there are 0,1 or 2  y values.
Sorry to read into your method wrong.

---

igloo myrtilles fourmis