Four Crossing Circles
The following engaging problem has been posted at the CTK Exchange by Bui Quang Tuan and its solution has been posted by Mariano Perez de la Cruz.
| Two circles (O1), (O2) intersect each other at A, B. Two circles (Oa), (Ob) centered at any point O: (Oa) passing A, (Ob) passing B.
Other than A: (Oa) intersects (O1), (O2) at A1, A2 respectively.
Other than B: (Ob) intersects (O1), (O2) at B1, B2 respectively.
Ra, Rb are radii of (Oa), (Ob) respectively. |
| Please prove: |
| 1. (Oa) cuts lines A1B1 and A2B2 at two equal segments. Similarly with (Ob). |
| 2. A1A2/B1B2 = Ra/Rb |
| |
|
Solution
|Activities|
|Contact|
|Front page|
|Contents|
|Geometry|
|Store|
Copyright © 1996-2012 Alexander Bogomolny
| Two circles (O1), (O2) intersect each other at A, B. Two circles (Oa), (Ob) centered at any point O: (Oa) passing A, (Ob) passing B.
Other than A: (Oa) intersects (O1), (O2) at A1, A2 respectively.
Other than B: (Ob) intersects (O1), (O2) at B1, B2 respectively.
Ra, Rb are radii of (Oa), (Ob) respectively. |
| Please prove: |
| 1. (Oa) cuts lines A1B1 and A2B2 at two equal segments. Similarly with (Ob). |
| 2. A1A2/B1B2 = Ra/Rb |
Observe that segment A1B1 is rotated around O, the common center of the two concentric circles (Oa), (Ob) into AB. AB in turn is rotated around O into A2B2. The product of two rotations is a rotation that maps A1B1 onto A2B2.
| |
|
The implication is twofold. First of all, the lines of A1B1 and A2B2 are at the same distance from O, implying that the chords cut off on these lines by the circle (Oa) are equal. The same of course holds for the circle (Ob). Secondly, the distance between a point and its image under a rotation is proportional to its distance from the center of rotation. Thus A1A2/B1B2 = Ra/Rb.
|Activities|
|Contact|
|Front page|
|Contents|
|Geometry|
|Store|
Copyright © 1996-2012 Alexander Bogomolny
|
