Let $AB$ be a diameter of a circle; let $t_1$ and $t_2$ be the tangents at $A$ and $B$, respectively; let $C$ be any point other than $A$ on $t_1$; and let $D_1D_2. E_1E_2$ be arcs on the circle determined by two lines through $C$. Prove that the lines $AD_1$ and $AD_2$ determine a segment on $t_2$ equal in length to that of the segment on $t_2$ determined by $AE_1$ and $AE_2.$
Problem
Source: IMO 1980 Austria-Poland, problem 9
Tags: symmetry, geometry, IMO Shortlist
grobber
09.05.2004 08:08
The lines $AD_1$ and $AE_1$ determine an involution on the tangent from $C$ to the circle (the tangent different from $CA$). This means that they also determina an ivnolution on $t_2$, and since it's obvious that the infinity point of $t_2$ corresponds to itself in this involution, it means that the involution is a symmetry wrt a point (a well-known rersult), and from here the conclusion follows (we find the midpoint of $D_1'E_1'$ to be the same as that of $D_2'E_2'$, where, in general, $X'=AX\cap t_2$).
thecmd999
18.07.2014 23:33
Let $X_1$, $X_2$, $Y_1$, $Y_2$ be the intersections of lines $AD_1$, $AD_2$, $AE_1$, $AE_2$ with $t_2$. Furthermore, let $CT$ be the other tangent from $C$ to the circle. Finally, let $M=AT\cap t_2$. Since quadrilateral $AE_1TE_1$ is harmonic, the perspectivity at $A$ onto $t_2$ yields $-1=(A, D_1;T, E_1)=(P_{\infty}, X_1;M, Y_1)$, where $P_{\infty}$ is the point at infinity on $t_2$. It follows that $M$ is the midpoint of segment $X_1Y_1$. Similarly, $M$ is the midpoint of segment $X_2Y_2$, so $X_1X_2=Y_1Y_2$ and the result follows. $\blacksquare$