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Problem 1206

AIME late

Geometry Difficulty 5.2 Prove it Croatian Mathematical Competitions · Croatia

Let $\triangle ABC$ be a triangle with an obtuse angle at vertex $B$ , let $D$ and $E$ be midpoints of the segments $\overline{AB}$ and $\overline{AC}$ respectively, let $F$ be a point on the segment $\overline{BC}$ such that $\angle BFE = 90^\circ$ , and let $G$ be a point on the segment $\overline{DE}$ such that $\angle BGE = 90^\circ$ . Prove that the points $A, F$ and $G$ are collinear if and only if $2|BF| = |CF|$ . (Matija Bašić)

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Obviously, $BFEG$ is a rectangle. The triangles $ADE$ and $ABC$ are similar with the coefficient of similarity $2$ . The points $A$ , $F$ and $G$ are collinear if and only if $|FB| : |GD| = 2 : 1$ .

Let $|BC| = a$ and $|BF| = x$ . Then $|ED| = \frac{1}{2}a$ , $|EG| = |FB| = x$ , $|GD| = |DE| - |EG| = \frac{1}{2}a - x$ , and the equality $|FB| = 2|GD|$ is equivalent to $x = 2\left(\frac{1}{2}a - x\right)$ , i.e. $a = 3x$ .

Therefore $|FB| = 2|GD|$ is equivalent to $|BC| = 3|BF|$ , i.e. $|CF| = 2|BF|$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.