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
StarLex1
17.11.2022 16:58
WLOG $M(0,0)$ with $r=1$ and $A(-1,0),B(1,0)$
$AC=AM=CM$
thus C could be $(-\frac{1}{2},\frac{\sqrt{3}}{2})$ or $(-\frac{1}{2},-\frac{\sqrt{3}}{2})$ just pick one
now pick $C (-\frac{1}{2},\frac{\sqrt{3}}{2})$
equation of line AC
$Y_{AC} =\sqrt{3}(x+1) $
thus $D= (\frac{1}{2},3\frac{\sqrt{3}}{2})$
$1+0+a*1+c=0$
$a+c=-1$
$1+(-\frac{1}{2})a+\frac{\sqrt{3}}{2}*b+c=0$
$b=\sqrt{3}{a}$
$\frac{1}{4}+\frac{27}{4}+\frac{1}{2}*a+(\frac{3\sqrt{3}}{2})*b+c=0$
$(a,b,c) = (\frac{-3}{2},-3\frac{\sqrt{3}}{2},\frac{1}{2})$
center of BCD wrt $x= -\frac{a}{2} =\frac{3}{4}$ wrt y $= \frac{3\sqrt{3}}{4}$
$r^2 = \frac{9}{4}-\frac{1}{2}=\frac{7}{4}$
$(x-\frac{3}{4})^2+(y-\frac{3\sqrt(3)}{4})^2=\frac{7}{4}$
since equation line AB,$y=0$
$(x-\frac{3}{4})^2=\frac{7}{4}-\frac{27}{16}$
meaning $E = (\frac{1}{2},0)$ since $E \ne B$
$D=(\frac{1}{2},\frac{3\sqrt{3}}{2})$
$Y_{bc} = \frac{(1-x)}{\sqrt{3}}$ at $x=\frac{1}{2}$
$F=(\frac{1}{2},\frac{1}{2\sqrt{3}})$
$Y_{AF}=Y_{BD}$
$\frac{(x+1)}{3\sqrt{3}} = 3\sqrt{3}(1-x)$
$X=(\frac{13}{14},\frac{3\sqrt{3}}{14})$
$\frac{BX}{XD}=\frac{\sqrt{\frac{1}{196}+\frac{27}{196}}}{\sqrt{\frac{36}{196}(28)}}=\frac{1}{6}$
ACGNmath
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 $\frac{BX}{XD}=\frac{5}{2}$.