IMO 2006 Shortlisted Problems

Let $K$ be the intersection point of lines $JC$ and $A_1{B}_1$. Obviously $JC\perp A_1{B}_1$ and since $A_1{B}_1\perp AB$, the lines $JK$ and $C_{1}D$ are parallel and equal. From the right triangle $B_{1}CJ$ we obtain $J{C}_1^2=J{B}_1^2=JC\cdot JK=JC\cdot C_{1}D$ from which we infer that $D{C}_1/C_{1}J=C_{1}J/JC$ and the right triangles $D{C}_{1}J$ and $C_{1}JC$ are similar. Hence $\angle C_{1}DJ=\angle J{C}_{1}C$, which implies that the lines $DJ$ and $C_{1}C$ are perpendicular, i.e the points $C_1,E,C$ are collinear.



Since $\angle C{A}_{1}J=\angle C{B}_{1}J=\angle CEJ=90^{\circ }$, points $A_1,B_1$ and $E$ lie on the circle of diameter $CJ$. Then $\angle DB{A}_1=\angle A_{1}CJ=\angle DE{A}_1$, which implies that quadrilateral $BE{A}_{1}D$ is cyclic; therefore $\angle A_{1}EB=90^{\circ }$.

Quadrilateral $ADE{B}_1$ is also cyclic because $\angle E{B}_{1}A=\angle EJC=\angle ED{C}_1$, therefore we obtain $\angle AE{B}_1=\angle ADB=90^{\circ }$.

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Alternative solution.

Consider the circles ${\omega }_1,{\omega }_2$ and ${\omega }_3$ of diameters $C_{1}D,A_{1}B$ and $A{B}_1$, respectively. Line segments $J{C}_1,J{B}_1$ and $J{A}_1$ are tangents to those circles and, due to the right angle at $D$, ${\omega }_2$ and ${\omega }_3$ pass through point $D$. Since $\angle C_{1}ED$ is a right angle, point $E$ lies on circle ${\omega }_1$, therefore $$J{C}_1^2=JD\cdot JE.$$ Since $J{A}_1=J{B}_1=J{C}_1$ are all radii of the excircle, we also have $$J{A}_1^2=JD\cdot JE\text{ and }J{B}_1^2=JD\cdot JE.$$ These equalities show that $E$ lies on circles ${\omega }_2$ and ${\omega }_3$ as well, so $\angle BE{A}_1=\angle AE{B}_1=90^{\circ }$.

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Alternative solution.

First note that $A_1{B}_1$ is perpendicular to the external angle bisector $CJ$ of $\angle BCA$ and parallel to the internal angle bisector of that angle. Therefore, $A_1{B}_1$ is perpendicular to $AB$ if and only if triangle $ABC$ is isosceles, $AC=BC$. In that case the external bisector $CJ$ is parallel to $AB$.

Triangles $ABC$ and $B_1{A}_{1}J$ are similar, as their corresponding sides are perpendicular. In particular, we have $\angle D{A}_{1}J=\angle C_{1}B{A}_1$; moreover, from cyclic deltoid $J{A}_{1}B{C}_1$, $$\angle C_1{A}_{1}J=\angle C_{1}BJ=\frac{1}{2}\angle C_{1}B{A}_1=\frac{1}{2}\angle D{A}_{1}J.$$ Therefore, $A_1{C}_1$ bisects angle $\angle D{A}_{1}J$.



In triangle $B_1{A}_{1}J$, line $J{C}_1$ is the external bisector at vertex $J$. The point $C_1$ is the intersection of two external angle bisectors (at $A_1$ and $J$ ) so $C_1$ is the centre of the excircle $\omega$, tangent to side $A_{1}J$, and to the extension of $B_1{A}_1$ at point $D$.

Now consider the similarity transform $\phi$ which moves $B_1$ to $A,A_1$ to $B$ and $J$ to $C$. This similarity can be decomposed into a rotation by $90^{\circ }$ around a certain point $O$ and a homothety from the same centre. This similarity moves point $C_1$ (the centre of excircle $\omega$ ) to $J$ and moves $D$ (the point of tangency) to $C_1$.

Since the rotation angle is $90^{\circ }$, we have $\angle XO\phi (X)=90^{\circ }$ for an arbitrary point $X\ne O$. For $X=D$ and $X=C_1$ we obtain $\angle DO{C}_1=\angle C_{1}OJ=90^{\circ }$. Therefore $O$ lies on line segment $DJ$ and $C_{1}O$ is perpendicular to $DJ$. This means that $O=E$.

For $X=A_1$ and $X=B_1$ we obtain $\angle A_{1}OB=\angle B_{1}OA=90^{\circ }$, i.e. $$\angle BE{A}_1=\angle AE{B}_1=90^{\circ }.$$