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Bulgaria geometry
Problem
The excircles of touch the sides , and at points , and , respectively. Let and be the incenter and the circumcenter of . Prove that if is a cyclic quadrilateral, then:
a) the points , and are collinear;
b) the points , and are collinear.
a) the points , and are collinear;
b) the points , and are collinear.
Solution
By Carnot's theorem, the perpendiculars from the points , and to the lines , and , respectively, have a common point . Then the quadrilateral is cyclic. Now it is easy to see that and hence is a diameter of the circumcircle of .
a) Denote by and the respective excenters of . Then Pappus' theorem for the triples of points and implies the desired result.
b) Let the incircle of touches the sides and at point and , respectively. Then the bisectors of and coincide with the bisectors of and , respectively, and pass through the midpoint of .
a) Denote by and the respective excenters of . Then Pappus' theorem for the triples of points and implies the desired result.
b) Let the incircle of touches the sides and at point and , respectively. Then the bisectors of and coincide with the bisectors of and , respectively, and pass through the midpoint of .
Techniques
Cyclic quadrilateralsTangentsPappus theoremTriangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle