Baltic Way 2023 · Problem 12

Geometry

Let $A B C$ be an acute triangle with $A B>A C$ . The internal angle bisector of $\angle B A C$ intersects $B C$ at $D$ . Let $O$ be the circumcentre of $A B C$ . Let $A O$ intersect the segment $B C$ at $E$ . Let $J$ be the incentre of $A E D$ . Prove that if $\angle A D O=45^{\circ}$ then $O J=J D$ .

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Topics

Angles and distances · Cyclic geometry · Triangles and centers

Solutions

Solution

Let $\alpha = \angle BAC$ , $\beta = \angle CBA$ , $\gamma = \angle ACB$ . We have

\begin{align*} \angle DJA &= 90^\circ + \frac{1}{2} \angle DEA = 90^\circ + \frac{1}{2} (\angle EBA + \angle BAE) \\ &= 90^\circ + \frac{1}{2} (\beta + 90^\circ - \gamma) = 135^\circ + \frac{\beta}{2} - \frac{\gamma}{2} \end{align*}

and

\begin{align*} \angle DOA &= 180^\circ - \angle OAD - \angle ADO = 180^\circ - (\angle OAC - \angle DAC) - 45^\circ \\ &= 135^\circ - \left(90^\circ - \beta - \frac{\alpha}{2}\right) = 135^\circ - \left(\frac{1}{2}(\alpha + \beta + \gamma) - \beta - \frac{\alpha}{2}\right) \\ &= 135^\circ + \frac{\beta}{2} - \frac{\gamma}{2} \end{align*}

Therefore, $\angle DJA = \angle DOA$ , hence quadrilateral $ADJO$ is cyclic. Since $AJ$ is the bisector of $\angle OAD$ , the arcs $OJ$ and $JD$ are equal. Hence $|OJ| = |JD|$ .

Contest context

Results from Baltic Way 2023

10 teams

Mean score: 2.0 / 5
Scores of 4 or 5: 3 / 10
Estonia: 3 / 5

Score distribution

All team scores

Team	Score
Germany	5 / 5
Sweden	0 / 5
Lithuania	0 / 5
Poland	5 / 5
Estonia	3 / 5
Latvia	1 / 5
Norway	0 / 5
Denmark	5 / 5
Finland	1 / 5
Iceland	0 / 5

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