Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.9 AIME, harder Prove it Estonia

Non-equilateral triangle $ABC$ has a $60°$ angle at vertex $A$ . Let the angle bisector drawn from vertex $A$ intersect the opposite side at point $D$ , and let $Q$ and $R$ be the feet of the altitudes drawn from vertices $B$ and $C$ , respectively. Prove that lines $AD$ , $BQ$ and $CR$ intersect in three distinct points that are vertices of an equilateral triangle.

Solution

If line $AD$ passed through the point of intersection of lines $BQ$ and $CR$ , the line segment $AD$ would be an altitude of triangle $ABC$ . As $AD$ is also the angle bisector, triangle $ABC$ would be isosceles with $AB = AC$ . As $\angle BAC = 60°$ , triangle $ABC$ would be equilateral, contradicting the assumption. Hence the lines $AD$ , $BQ$ , and $CR$ meet in three distinct points. Fig. 15

Fig. 15

By assumptions, $\angle QAD = \angle RAD = 30°$ and $\angle AQB = \angle ARC = 90°$ (Fig. 15). Hence $AD$ and $BQ$ intersect at angle $60°$ , as well as $AD$ and $CR$ . Thus two angles of the triangle whose vertices are the three intersection points of lines $AD$ , $BQ$ and $CR$ have size $60°$ . Such a triangle is equilateral.

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