Let $k$ be a circle with centre $M$ and let $AB$ be a diameter of $k$. Furthermore, let $C$ be a point on $k$ such that $AC = AM$. Let $D$ be the point on the line $AC$ such that $CD = AB$ and $C$ lies between $A$ and $D$. Let $E$ be the second intersection of the circumcircle of $BCD$ with line $AB$ and $F$ be the intersection of the lines $ED$ and $BC$. The line $AF$ cuts the segment $BD$ in $X$. Determine the ratio $BX/XD$.
Problem
Source: Switzerland - 2022 Swiss Final Round p1
Tags: ratio, geometry
17.11.2022 16:58
17.11.2022 17:22
[asy][asy] size(10cm); pair M = (0,0); pair A = (-1,0); pair B = (1,0); pair C = dir(120); pair D = 2*A-C; pair E = foot(D,A,B); pair F = extension(D,E,C,B); pair X = foot(F,B,D); draw(F--D--B--F--cycle); draw(B--E); draw(D--C); draw(F--X); draw(circumcircle(B,C,D)); draw(unitcircle); dot("$A$",A,dir(135)); dot("$B$",B,dir(B)); dot("$C$",C,dir(100)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$X$",X,dir(270)); dot("$M$",M,dir(45)); [/asy][/asy] Note that $AM=MC=CA$, thus $\triangle ACM$ is equilateral. Thus $\angle CBA=30^{\circ}$ and $\angle DFB=60^{\circ}$. Next, note that since $DECB$ is cyclic and $\angle DCB=90^{\circ}$, we have $\angle DEB=90^{\circ}$. This means that $BE$ and $DC$ are heights in $\triangle FDB$, and $A$ is the orthocenter. Thus $X$ is the foot of the perpendicular from $F$ to $BD$, and lies on $(ABC)$. Scale the figure such that $AD=AC=1$. Then we have $AB=2$, and $CB=\sqrt{3}$. Let $FA=k$. Since $FA\cdot AX=DA\cdot AC=1$, we have $AX=\frac{1}{k}$. Then we have \[FA\cdot FX=FC\cdot FB\]\[k\left(k+\frac{1}{k}\right)=\sqrt{k^-1}(\sqrt{k^2-1}+\sqrt{3})\]\[k^2+1=k^2-1+\sqrt{3(k^2-1)}\]\[2=\sqrt{3(k^2-1)}\]\[k=\sqrt{\frac{7}{3}}\]Thus, $AX=\sqrt{\frac{3}{7}}$. This means that $DX=\sqrt{\frac{4}{7}}$ and $BX=\sqrt{\frac{25}{7}}$, which means that $\frac{BX}{XD}=\frac{5}{2}$.