Selected Problems from Open Contests

If $CD$ is the diameter of the circumcircle of triangle $ABC$, then $A_1=A$ and $B_1=B$ and the statement holds. Assume that $CD$ is not the diameter (Fig. 3). Then $A_1\ne A$ and $B_1\ne B$. The point $A_1$ lies on the ray $AD$ if and only if the point $B_1$ does not lie on the ray $BD$ (depending on which side of the diameter through point $C$ point $D$ is located). Thus, $\angle CA{A}_1=$ $\angle CB{B}_1$ (because the sum of opposite angles of a cyclic quadrilateral $ADBC$ is $180^{\circ }$). Thus, the right-angled triangles $A{A}_{1}C$ and $B{B}_{1}C$ are similar. From $\angle AC{A}_1=\angle BC{B}_1$ we see that $\angle ACB=\angle A_{1}C{B}_1$. This together with $\frac{|AC|}{|BC|}=\frac{|A_{1}C|}{|B_{1}C|}$ gives that $ACB$ and $A_{1}C{B}_1$ are similar. As $|AC|>|A_{1}C|$, we conclude that $|AB|>|A_1{B}_1|$.

Solution 2:

Since the angles $C{A}_{1}D$ and $C{B}_{1}D$ are right angles, the points $C,A_1,D$, and $B_1$ form a cyclic quadrilateral and thus $\angle CAB=\angle CDB=\angle C{A}_1{B}_1$. Similarly, $\angle CBA=\angle C{B}_1{A}_1$. Therefore the triangles $ABC$ and $A_1{B}_{1}C$ are similar. As $|CA|\ge |C{A}_1|$, we deduce that $|AB|\ge |A_1{B}_1|$.

Fig. 3

Solution 3:

The radius $R$ of the circumcircle of the quadrilateral $CADB$ is at least as large as the radius $R_1$ of the circumcircle of the quadrilateral $C{A}_{1}D{B}_1$ because $CD$ is a chord in the first one and a diameter in the second one. The sine law in triangles $ADB$ and $A_{1}D{B}_1$ gives $|AB|=2R\sin \angle D$ and $|A_1{B}_1|=2R_1\sin \angle D$. As $R\ge R_1$, we deduce $|AB|\ge |A_1{B}_1|$.