2017 IMO Problems/Problem 4
Contents
Problem
Let and be different points on a circle such that is not a diameter. Let be the tangent line to at . Point is such that is the midpoint of the line segment . Point is chosen on the shorter arc of so that the circumcircle of triangle intersects at two distinct points. Let be the common point of and that is closer to . Line meets again at . Prove that the line is tangent to .
Solution
We construct inversion which maps into the circle and into Than we prove that is tangent to
Quadrangle is cyclic
Quadrangle is cyclic
We construct circle centered at which maps into
Let Inversion with respect to swap and maps into
Let be the center of
Inversion with respect to maps into . belong circle is the image of . Let be the center of
is the image of at this inversion, is tangent line to at so
is image K at this inversion is parallelogram.
is the midpoint of is the center of symmetry of is symmetrical to with respect to is symmetrical to with respect to is symmetrycal with respect to
lies on and on is tangent to line is tangent to
vladimir.shelomovskii@gmail.com, vvsss
Solution 2
We use the tangent-chord theorem: the angle formed between a chord and a tangent line to a circle is equal to the inscribed angle on the other side of the chord.
Quadrangle is cyclic
Quadrangle is cyclic
(One can use Reim’s theorem – it is shorter way.)
Let be symmetric to with respect to is parallelogram. is cyclic.
Inscribed angle of is equal to angle between and chord
is tangent to by the inverse of tangent-chord theorem.
vladimir.shelomovskii@gmail.com, vvsss
See Also
2017 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |