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Romanian Mathematical Olympiad

Geometry Difficulty 6.2 National Olympiad Prove it Romania

Let $\triangle ABC$ be a triangle in which $\angle ABC = 75^\circ$ and $\angle BAC = 45^\circ$ . One considers the points $F$ , $X$ and $Y$ such that $F$ is the projection of $B$ on $AC$ , $\overline{CX} = \frac{1}{2}\overline{BC}$ and $\overline{BY} = \overline{FX} + \overline{FA}$ . Prove that the centroid of triangle $ABX$ lies on the line segment $YF$ .

Solution

We will prove that $YF$ is the Euler line of the triangle $ABX$ , hence it contains the centroid of that triangle.

Because $BFA$ is a right isosceles triangle, it follows that $BF = AF$ . In the right triangle $BFC$ we have $\angle CBF = 30^\circ$ , hence $FC = \frac{BC}{2}$ , thus $FC = CX$ , and a short computation shows that $\angle BXF = 30^\circ$ . We deduce that $BF = FX$ , hence $AF = BF = FX$ , making $F$ the circumcenter of $ABX$ .

Let $Y'$ be the orthocenter of $ABX$ . Applying Sylvester's relation yields $\overline{FY'} = \overline{FA} + \overline{FB} + \overline{FX} = \overline{FB} + \overline{BY} = \overline{FY}$ . It follows that $Y' = Y$ , hence $Y$ is the orthocenter of $ABX$ . We conclude that $FY$ is the Euler line of $ABX$ and since the centroid lies between the circumcenter and the orthocenter, we obtain the desired result.

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