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Radical Axis and Center, an ApplicationThe applet below presents a problem from Mathematical Miniatures, an exciting book by MAA. Let points C and D lie on a semicircle with diameter AB. Points E, F, G are the midpoints of chords AC, CD, and DB, respectively. Let the perpendicular from E to AF intersect the (vertical) tangent to the semicircle at A in M, and the perpendicular from G to BF intersect the tangent at B in N. Then MN is parallel to CD.
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Copyright © 1996-2009 Alexander Bogomolny
Radical Axis and Center, an Application
The quadrilateral OEFG is the Varignon parallelogram of the quadrilateral ACDB. Its diagonals OF and EG bisect each other in, say, O3. Denote the original (semi)circle as c and that with diameter EG as c3. I want to show that MN serves as the radical axis of the two circles. Since, the center line OO3 is perpendicular to the radical axis of the two circles, and coincides with OF, which is perpendicular to CD, this would imply that MN is indeed parallel to CD. Introduce circles c1 and c2 with diameters AO and OB, respectively. E is the midpoint of the chord AC in c, so that EO is perpendicular to AC. Thus, the angle AEO is right, and E lies on c1. Similarly, G lies on c2. O1O3 is the midline in
Copyright © 1996-2009 Alexander Bogomolny |
AOF and is therefore parallel to AF. By the construction, the center line O1O3 is then perpendicular to EM. But so is the radical axis of c1 and c3. Since both pass through the points of intersection (E, in particular) of c1 and c3, we may claim that EM is the radical axis of c1 and c3. On the other hand, the radical axis of c and c1 is their common tangent at A. The two axes intersect at the | 34222007 |  |
