Estonian Mathematical Olympiad

Solution 1:

Let $\angle BCA=\gamma$ and let $M$ be the point of intersection of lines $OJ$ and $AB$ (Figures 10 and 11). As $OA=OB$ and $OM$ bisects the angle $AOB$, we have $OM\perp AB$. We also have $\angle AOM=\angle AOB=\angle ACB=\gamma$, because $O$ lies inside the triangle $ABC$. As $\angle AMO=90^{\circ }=\angle ADC$, triangles $AMO$ and $ADC$ are similar. Hence also $\angle MAO=\angle DAC=90^{\circ }-\gamma$, implying $$\angle MAJ=\angle JAO=\angle DAE=\angle EAC=\frac{1}{2}(90^{\circ }-\gamma ).(6)$$ Hence triangles $AJO$ and $AEC$ are similar, too. This implies the similarity of triangles $AJE$ and $AOC$ (spiral similarity). As $OA=OC$, the latter similarity implies $JA=JE$. By symmetry in the isosceles triangle $AOB$, we obtain $JA=JB$. Consequently, $JB=JE$. Hence the triangle $JBE$ is isosceles.

Fig. 10 Fig. 11

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Alternative solution.

Solution 2:

Let $\angle CAB=\alpha$, $\angle ABC=\beta$ and $\angle BCA=\gamma$. We start by proving the equalities (6) as in Solution 1. Next, let lines $OJ$ and $BC$ meet in $F$, whereas let lines $AJ$ and $BC$ meet in $X$ (Figures 12 and 13 depict situations that differ by the order of points $E$ and $F$). We show that points $A,J,E,F$ are concyclic. From equalities (6) we get $$\begin{array}{rl}\angle JAE& =\angle BAC-\angle BAJ-\angle EAC\\ & =\alpha -\frac{1}{2}(90^{\circ }-\gamma )-\frac{1}{2}(90^{\circ }-\gamma )\\ & =\alpha -(90^{\circ }-\gamma )=\alpha +\gamma -90^{\circ }=180^{\circ }-\beta -90^{\circ }=90^{\circ }-\beta .\end{array}$$ As $OA=OB$, the line $OJ$ is the perpendicular bisector of the line segment $AB$. Thus $$\angle XFJ=\angle BFJ=90^{\circ }-\angle ABF=90^{\circ }-\beta .$$

Fig. 12 Fig. 13 Altogether, we have $\angle XAE=\angle JAE=90^{\circ }-\beta =\angle XFJ$, implying that $A,J,E,F$ are concyclic.

Therefore $\angle JEA=\angle JFA=\angle JFB=\angle JFX=\angle XAE=\angle JAE$, implying $JA=JE$. But $JA=JB$ as $J$ lies on the perpendicular bisector of the line segment $AB$. Consequently, $JB=JE$. Hence the triangle $JBE$ is isosceles.

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Alternative solution.

Solution 3:

We use the known fact that lines drawn from a vertex of a triangle to its orthocentre and its circumcentre are isogonals which implies that the angles between these lines and the respective sides are equal. As $AD$ and $AO$ are the altitudes and the circumradius of the triangle $ABC$, both drawn from vertex $A$, we have $\angle BAO=\angle CAD$. But $AE$ bisects the angle $CAD$ and $J$ is the point of intersection of angle bisectors of the triangle $ABO$, hence $\angle BAJ=\angle CAD=\angle EAD$. This shows that rays $AJ$ and $AD$ are isogonals in the triangle $ABE$. As $AD$ is also an altitude of the triangle $ABE$, the line $AJ$ must pass through the circumcentre of the triangle $ABE$. But the circumcentre of the triangle $ABE$ must also lie on the perpendicular bisector of the side $AB$. As $OA=OB$, the bisector of the angle $AOB$ coincides with the perpendicular bisector of the side $AB$. Hence the circumcentre of the triangle $ABE$ is the point $J$ of intersection of lines $AJ$ and $OJ$. Consequently, $JB=JE$, implying that the triangle $JBE$ is isosceles.

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Alternative solution.

Solution 4:

As $OA=OB$, the bisector of the angle $AOB$ coincides with the perpendicular bisector of the side $AB$. Hence the point $J$ lies on the perpendicular bisector of the side $AB$, implying that points $A$ and $B$ lie on some circle $\omega$ with centre $J$. We show that $\angle AJB=2\angle AEB$. Indeed, $$\begin{array}{rl}\angle AEB& =\angle AED=90^{\circ }-\angle EAD\\ & =90^{\circ }-\frac{\angle CAD}{2}=90^{\circ }-\frac{90^{\circ }-\angle ACD}{2}=90^{\circ }+\frac{\angle ACD}{2}.\end{array}$$ As $JA=JB$ implies $\angle JAB=\angle JBA$, we obtain $$\begin{array}{rl}\angle AJB& =180^{\circ }-2\angle JAB=180^{\circ }-\angle OAB\\ & =180^{\circ }-\frac{180^{\circ }-\angle AOB}{2}=90^{\circ }+\frac{\angle AOB}{2}=90^{\circ }+\angle ACB=2\angle AEB.\end{array}$$

Hence $\omega$ is the circumcircle of the triangle $ABE$. Consequently, $JB=JE$, implying that the triangle $JBE$ is isosceles.