BMO 2010 Shortlist

Let the lines $DE$ and $CA$ meet at point $L$. We will prove that the line $BK$ passes through $L$ (see figure 1). The circle $c(O,R)$ is homothetic to the circle $c_1(K,R_1)$ with respect to homothety with center $A$ and ratio $m=\frac{R}{R_1}$, say $H(A,\frac{R}{R_1})$. The extension of $CD$ meets the circle ($c$) at a point $D_1$ homothetic of $D$. The extension of $CE$ meets the circle ($c$) at a point $E_1$ homothetic of $E$. Therefore the line segment $C{E}_1$ is homothetic of the line segment $CE$. So, if the line $AC$ intersects $D_1{E}_1$ at the point $L_1$, then $L_1$ will be homothetic of $L$. Since $O$ is homothetic of $K$, we conclude that $$O{L}_1\parallel KL.(1)$$ We will prove that $$O{L}_1\parallel BL.(1)$$ Since $AD$ and $AE$ are tangents from $A$ to the circle ($c_1$), then $AK$ is the perpendicular bisector of the segment $DE$. Let $M$ be the intersection point of the lines $AK$ and $DE$, such that the extension of $CM$ intersects $D_1{E}_1$ at $M_1$ and the circle ($c$) at point $M_2$. Then $M_1$ will be the middle point of the segment $D_1{E}_1$ (because of the homothety). We assert that $CA$ is the symmedian of the triangle $CDE$ which corresponds to the vertex $C$. According to Steiner's theorem to symmedians, it is enough to prove that $$\frac{DL}{LE}=\frac{C{D}^2}{C{E}^2}.(3)$$ For proving the relation (3) we use the areas ratio: $$\frac{\sigma (CDL)}{\sigma (CEL)}=\frac{DL}{LE}=\frac{\sigma (DAL)}{\sigma (EAL)}=\frac{\sigma (CDL)+\sigma (DAL)}{\sigma (CEL)+\sigma (EAL)}=\frac{\sigma (CAD)}{\sigma (CAE)}.(4)$$ Since the angles $ADE$ and $AED$ are the angles between tangents and chord we have $\angle ADE=\angle AED=\angle DCE$ and therefore $$\angle CDA=\angle CDE+\angle ADE=\angle CDE+\angle DCE=180^{\circ }-\angle CED,$$ $$\angle CEA=\angle CED+\angle AED=\angle CED+\angle DCE=180^{\circ }-\angle CDE.$$ From (4) we obtain $$\frac{DL}{LE}=\frac{\sigma (CAD)}{\sigma (CAE)}=\frac{CD\cdot \sin (180^{\circ }-\angle CED)}{CE\cdot \sin (180^{\circ }-\angle CDE)}=\frac{CD\cdot \sin (\angle CED)}{CE\cdot \sin (\angle CDE)}=\frac{C{D}^2}{C{E}^2}.$$ Figure 1 So, the relation (3) is proved and $CA$ is the symmedian of the triangle $CDE$ which corresponds to the vertex $C$. Hence $\angle D_{1}CA=\angle E_{1}C{M}_2$ and the quadrilateral $A{D}_1{E}_1{M}_2$ is an isosceles trapezium. The line $O{M}_1$ is perpendicular to $D_1{E}_1$ and intersects $A{M}_2$ at the middle $N$. In the triangle $AM{M}_2$ we have that $N$ is the middle of the side $A{M}_2$ and $N{M}_1\parallel AM$. Hence $M_1$ is the middle of $M{M}_2$ and therefore $D_1{E}_1$ is the mid-parallel of $A{M}_2$ and $DE$. Since $L_1$ belongs to $D_1{E}_1$, it will be the middle of $AL$. In the triangle $ALB$ $O{L}_1$ is the mid-parallel to $BL$. Hence $O{L}_1\parallel BL$ and the straight lines $AC$, $BK$ and $DE$ are concurrent. $\square$