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.
Wednesday, February 7, 2007
Challenge Problem for 2-7-07
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7 comments:
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
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.
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!
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?
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)
OK,
So now I have to analyze for my goof(s). The kids' approaches were quite different from mine.
That's a cool problem; what a neat relationship!
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