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.
What if applet does not run? |
References
- S. Savchev, T. Andreescu, Mathematical Miniatures, MAA, 2003, pp. 114-115
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny
Radical Axis and Center, an Application
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
- How to Construct a Radical Axis
- A Property of the Line IO: A Proof From The Book
- Cherchez le quadrilatere cyclique II
- Circles On Cevians
- Circles And Parallels
- Circles through the Orthocenter
- Coaxal Circles Theorem
- Isosceles on the Sides of a Triangle
- Properties of the Circle of Similitude
- Six Concyclic Points
- Radical Axis and Center, an Application
- Radical axis of two circles
- Radical Axis of Circles Inscribed in a Circular Segment
- Radical Center
- Radical center of three circles
- Steiner's porism
- Stereographic Projection and Inversion
- Stereographic Projection and Radical Axes
- Tangent as a Radical Axis
- Two Circles on a Side of a Triangle
- Pinning Butterfly on Radical Axes
- Two Lines - Two Circles
- Two Triples of Concurrent Circles
- Circle Centers on Radical Axes
- Collinearity with the Orthocenter
- Six Circles with Concurrent Pairwise Radical Axes
- Six Concyclic Points on Sides of a Triangle
- Line Through a Center of Similarity
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny
71492807