 # Parabolic sector Swlabr Posers and Puzzles 03 Nov '08 17:14
1. 03 Nov '08 17:141 edit
This may be hard to explain, and may in fact be impossible. I have been trying to work it out for a while, and failing badly... Although I'm sure it's not - my notes say the answer is a ratio of 3:1, but there is a simple counter example, and I think I may have copied them down wrongly... Anyway:

You have a circle, centred at the origin; draw a straight line, A, through it that has positive gradient. Draw the tangent, B, to the circle that is parallel to this line A (the tangent "closest" to your line). Connect these two lines with another line, C, that is parallel to the tangent to the base of the circle (this line would have equation x=a, 'a' a constant).

Connect the two places where line A cuts the circle to the place where line B meets line C. Now, again take the two points where line A meets circle. From the top point draw a vertical line, D, and from the bottom one a line, E, which is perpendicular to D. Shear line D off where it meets the circle and where it meets line E, and shear line E off where it meets line D. You should then have a circle with 2 triangles inside it, one of whish is right-angled. Also, line C should cut line D in half.

Call the point where lines A and D meet a, and where C and B meet b. Join these two points with a line, and continue this line until it cuts line E at a point, c. Take the length of line E between where it cuts line D and point c, and then mark this distance further along the line from c. Call this point d. Then the distance from d to where D meets E is equal to twice the distance between where D meets E and c.

Call the point where lines A and E meet s, and where D and E meet t.

Then take point a and join it to point d. How are triangles abs and atd connected?

I hope that is forumlated correctly and clearly...