NMO Selection Tests for the Balkan and International Mathematical Olympiads

Let the circle $ATX$ and the line $A_2{B}_1$ meet again at the point $Y$, and let the circle $BTX$ and the line $A_1{B}_2$ meet again at the point $Z$. Notice that the lines $AB,AY$ and $AZ$ are all parallel to the line $A_1{B}_1$ to deduce that the points $A,B,Y$ and $Z$ are collinear.

Next, consider the cyclic quadrangles $A_1{B}_1{A}_2{B}_2$, $A_1{B}_1{B}_{2}T$ and $BT{B}_{2}Z$ to get successively $\angle B_2{A}_{2}Y=\angle B_1{A}_1{B}_2=\angle B_{1}T{B}_2=\angle BZ{B}_2$, and infer thereby that the points $A_2,B_2,Y$ and $Z$ are co-cyclic. It then follows that the lines $TX,A_1{B}_2$ and $A_2{B}_1$ are the radical lines of the pairs of circles $ATX$ and $BTX,BTX$ and $A_2{B}_{2}YZ$, and $A_2{B}_{2}YZ$ and $ATX$, respectively, whence the conclusion.

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Alternative solution.

Apply an inversion of pole $T$ to rephrase the statement as follows Let ${\delta }_1$ and ${\delta }_2$ be two parallel lines, let $T$ be a point in the complement of ${\delta }_1\cup {\delta }_2$, and let ${\ell }_1$ and ${\ell }_2$ be two lines through $T$. The lines ${\ell }_1$ and ${\ell }_2$ meet ${\delta }_1$ at points $A$ and $B$, respectively, and ${\delta }_2$ at points $A_1$ and $B_1$, respectively. Let further $X$ be a point in the complement of ${\delta }_1\cup {\delta }_2\cup {\ell }_1\cup {\ell }_2$. The lines $AX$ and $BX$ meet ${\delta }_2$ at points $A_2$ and $B_2$, respectively. Prove that the circles $T{A}_1{B}_2$ and $T{A}_2{B}_1$ meet again at a point on the line $TX$.

In other words, we must prove that the line $TX$ is the radical line of the circles $T{A}_1{B}_2$ and $T{A}_2{B}_1$. It is sufficient to show that the point $Y$ where the line $TX$ meets ${\delta }_2$ is of equal powers with respect to the two circles. To this end, apply Menelaus' theorem to triangles $A{A}_1{A}_2$ and $B{B}_1{B}_2$ and transversal $TXY$, to get, upon multiplication and suitable rearrangement of factors, $$Y{A}_1\cdot Y{B}_2=(Y{A}_2\cdot Y{B}_1)\left(\frac{TB}{T{B}_1}\cdot \frac{T{A}_1}{TA}\right)\left(\frac{XA}{X{A}_2}\cdot \frac{X{B}_2}{XB}\right).$$ Finally, notice that the last two products in the parentheses above both equal 1, by similarity of triangles $TAB$ and $T{A}_1{B}_1$, and $XAB$ and $X{A}_2{B}_2$, to conclude that $Y$ is indeed of equal powers with respect to the circles $T{A}_1{B}_2$ and $T{A}_2{B}_1$.