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Baltic Way 1992 · Problem 19

Geometry

Let CC be a circle in the plane. Let C1C_{1} and C2C_{2} be non-intersecting circles touching CC internally at points AA and BB respectively. Let tt be a common tangent of C1C_{1} and C2C_{2}, touching them at points DD and EE respectively, such that both C1C_{1} and C2C_{2} are on the same side of tt. Let FF be the point of intersection of ADA D and BEB E. Show that FF lies on CC.

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Topics

Circles and tangency · Transformations

Solutions

Solution

Solution:

Let F1F_{1} be the second intersection point of the line ADA D and the circle CC (see Figure 3). Consider the homothety with centre AA which maps DD onto F1F_{1}. This homothety maps the circle C1C_{1} onto CC and the tangent line tt of C1C_{1} onto the tangent line of the circle CC at F1F_{1}. Let us do the same with the circle C2C_{2} and the line BEB E: let F2F_{2} be their intersection point and consider the homothety with centre BB, mapping EE onto F2F_{2}, C2C_{2} onto CC and tt onto the tangent of CC at point F2F_{2}. Since the tangents of CC at F1F_{1} and F2F_{2} are both parallel to tt, they must coincide, and so must the points F1F_{1} and F2F_{2}.

Contest context

Results from Baltic Way 1992

8 teams

Mean score
3.9 / 5
Scores of 4 or 5
6 / 8
Estonia
0 / 5

Score distribution

01
11
20
30
40
56
All team scores
TeamScore
Denmark1 / 5
St. Petersburg5 / 5
Poland5 / 5
Latvia5 / 5
Iceland5 / 5
Lithuania5 / 5
Estonia0 / 5
Sweden5 / 5