Library / Austrian Mathematical Olympiad /41 of 54

Austrian Mathematical Olympiad

Geometry Difficulty 6.5 National Olympiad Prove it Austria

Let $ABC$ be an isosceles triangle with $AC = BC$ and $\angle ACB < 60^\circ$ . We denote the incenter and circumcenter by $I$ and $O$ , respectively. The circumcircle of triangle $BIO$ intersects the leg $BC$ also at point $D \neq B$ .

a. Prove that the lines $AC$ and $DI$ are parallel.

b. Prove that the lines $OD$ and $IB$ are mutually perpendicular.

Solution

Note that the condition $\angle ACB < 60^\circ$ guarantees that $O$ lies between $I$ and $C$ .

a.
We denote the angles of triangle $ABC$ by $\alpha = \angle BAC$ , $\beta = \angle ABC$ and $\gamma = \angle ACB$ . Let $K$ and $k$ be the circumcircles of $ABC$ and $BIO$ , respectively. The inscribed angle theorem for circle $K$ yields: $\angle BOC = 2\alpha$ . Therefore we have $\angle IOB = 180^\circ - 2\alpha$ and because of $\alpha = \beta$ we obtain $\angle IOB = \gamma$ . Furthermore the inscribed angle theorem for circle $k$ gives $\angle IDB = \gamma$ , whence finally $ID \parallel AC$ .

b.
We denote the point of intersection of lines $OD$ and $IB$ by $F$ and the midpoint of $AB$ by $G$ . Since $IODB$ is cyclic, we have $\angle IOD = 180^\circ - \beta/2$ , that is $\angle DOC = \beta/2$ or equivalently $\angle FOI = \beta/2$ . Furthermore $\angle GIB = 90^\circ - \beta/2$ implies $\angle OIF = 90^\circ - \beta/2$ . Therefore $\angle IFO = 90^\circ$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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