Radical Axis and Center, an Application

The 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.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Proof.

References

  1. S. Savchev, T. Andreescu, Mathematical Miniatures, MAA, 2003, pp. 114-115

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Copyright © 1996-2018 Alexander Bogomolny

Radical Axis and Center, an Application


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

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 ΔAOF 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 radical center of c, c1 and c3, which is M. Being the radical center of the circle triad c, c1 and c3, M lies on the radical axis of c and c3. In a similar fashion, N is the radical center of the triad c, c2 and c3 and, as M, also lies on the radical axis of c and c3. We see that MN is in fact the radical axis of c and c3, which concludes the proof.

Radical Axis and Radical Center

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Copyright © 1996-2018 Alexander Bogomolny

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