Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it North Macedonia

Let $ABC$ be an isosceles triangle with $\overline{AB} = \overline{AC}$ . Let $D$ be the midpoint of $BC$ , $M$ the midpoint of $AD$ and $N$ the projection of $D$ to $BM$ . Prove that $\angle ANC = 90^\circ$ .

Solution

Let $S$ be the point so that $ABCD$ is a parallelogram. Then $ADCS$ is a rectangle and $R$ is the intersection point of the diagonals $AC$ and $DS$ . The point $N$ lies on the diagonal $BS$ of the parallelogram $ABDS$ from where we obtain that $SND$ is a right triangle. The point $R$ is a circumcenter for the triangle $SND$ , from where $\overline{NR} = \frac{1}{2}\overline{DS} = \frac{1}{2}\overline{AC}$ . The angle $\angle ANC = 90^\circ$ i.e. $ANC$ is a right triangle, because $\overline{RA} = \overline{RC} = \overline{RN}$ .

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