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PrintSaudi Arabia Mathematical Competitions 2012
Saudi Arabia 2012 geometry
Problem
Let be a triangle, the excenter opposite to , and its reflection across . Prove that is parallel to the Euler line of triangle .
Solution
Let be the incenter of , the orthocenter of , and the midpoint of the segment .
We have so is the circumcenter of triangle .
Moreover, (both are perpendicular to ) and, similarly, , hence is a parallelogram.
Let denote the projection of onto and the midpoint of segment . We have , hence Therefore The relation (1) proves that is parallel to . But , hence , and we are done.
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Alternative solution.
We will use the notation in the previous solution. The quadrilateral is cyclic, inscribed in a circle of diameter , which implies is the midpoint of segment . Also, we have that is a parallelogram.
We want to prove that We have Hence and The relation (3) implies We have a relation which is obviously true. It follows that and therefore .
We have so is the circumcenter of triangle .
Moreover, (both are perpendicular to ) and, similarly, , hence is a parallelogram.
Let denote the projection of onto and the midpoint of segment . We have , hence Therefore The relation (1) proves that is parallel to . But , hence , and we are done.
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Alternative solution.
We will use the notation in the previous solution. The quadrilateral is cyclic, inscribed in a circle of diameter , which implies is the midpoint of segment . Also, we have that is a parallelogram.
We want to prove that We have Hence and The relation (3) implies We have a relation which is obviously true. It follows that and therefore .
Techniques
Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circleCyclic quadrilateralsTriangle trigonometryAngle chasingConstructions and loci