## Wednesday, February 7, 2007

### Challenge Problem for 2-7-07

Day 8 - still awaiting a response from the National Math Panel...

Pls read the comments re the problems from 2-6-07. There are hints for #1 and a good discussion about #3.

Something different today. This is another one of those weekly online challenges I gave last year just before the AMC Contest. Students found it difficult but those who persisted got it. Perhaps that's the best reason to give these challenges -- to teach persistence, a quality that distinguishes some of the best researchers and problem solvers from the rest. Remember to click on the image to magnify it if it's too small.

Totally_clueless said...

Really nice problem.

I found it easier to do (b) first, and then find the intersection with the straight line that passes through (0,3) with slope -1/2 to find the two centers for (a).

For (c), I don't really know, but I should get the answer if I focus on the problem long enough :-).

I might also try looking for a better sense of directrix.

TC

Eric Jablow said...

To consider part (c) without equations, draw the line y = −1 under the diagram; call this L. Then, PQ = 1 + the radius of the second circle. The distance from Q to L is the radius of the second circle plus 1; thus these are equal. Therefore, the locus is the parabola with focus P and directrix L. Now, write the equation.

The tricky part is, of course, getting the idea to draw line L, and that requires that one recognizes the curve as a parabola.

Dave Marain said...

Folks, I will post 2 of the students' solutions on Thu. I think you will find their analysis interesting. Most struggled with this question for some time. I didn't receive any responses until near the 48-hour deadline! Are you surprised or do you think they had better things to do that try this!

tc--
are you trying to prove you're a worse punster than me! well, you do have a habit of going off on a tangent but actually i think you've gotten to the 'hearth' of the problem!

eric--
nice analysis!
Originally i envisioned this as a classic locus problem, free of all coordinates:
What is the locus of points of the centers of all circles externally tangent to a given circle and tangent to a given line which does not intersect the circle. There's nothing special about the x-axis in the given problem other than setting up the equation. Any other line outside the circle would lead to a parabola!

jonathan said...

Without the slope condition:
The distance between the centers (the sum of the radii) should be 1 more than the x-coordinate of the unknown center
x^2 + (y-3)^2 = (y-1)^2
x^2 + y^2 - 6y + 9 = y^2 - 2y + 1
x^2 - 4y + 8 = 0
y = (1/4)x^2 + 2

I don't know how to analyze the curve. Is the point the focus, and y = -1 the directrix?

jonathan said...

The distance between the centers (the sum of the radii) should be 1 more than the y-coordinate of the unknown center

(y-coordinate, not x-coordinate. Oops)

Jonathan said...

OK,

So now I have to analyze for my goof(s). The kids' approaches were quite different from mine.

Darmok said...

That's a cool problem; what a neat relationship!