Using Ptolemy's Theorem to find length ratio

Question

In this figure, $X, Y$ are tangent points and $\frac{DX}{EX} = \frac{8}{3} , \frac{EY}{DY} = 4 , \frac{AC}{AB} = \frac{5 }{4} . $

Then, what is $ \frac{BC}{AX}$ ?

System of equations from the condition are so cumbersome. How can I extract the core relation?

Answer 1

WLOG $DE=1$. Note that $B$ and $C$ are the centres of the Apollonius circles of $Y$ and $X$ with respect to $DE$ respectively. A proof of this property is given below:

Let $P_1$ and $P_2$ be the internal and external points on $DE$ such that $\frac{P_iD}{P_iE} = k > 1$, for some fixed real $k$. Let $M$ be the midpoint of $P_1P_2$. Then, we have $DP_1 = \frac{k}{k+1}, DP_2 = \frac{k}{k-1}$, so $P_1P_2 = \frac{2k}{k^2-1}$.

Thus, $MP_1 = MP_2 = \frac{k}{k^2-1}$, and $DM = \frac{k^2}{k^2-1}, EM = \frac{1}{k^2-1}$. This implies $MD \cdot ME = \frac{k^2}{(k^2-1)^2} = MP_1^2$.

From this, if an arbitrary circle $\Gamma$ passes through $DE$, then the intersection of $\Gamma$ with the circle centred at $M$ through $P_1$ and $P_2$ (call it $Q$) will also be the point such that $MQ$ is tangent to $\Gamma$.

Thus, reverse reconstructing, we get $C$ is the centre of the Apollonius circle of $X$, and similar for $B$.

Now, from above, we have the property that $k = \frac{8}{3}$ for $X$, so we get $CX = \frac{\frac{8}{3}}{\left ( \frac{8}{3} \right )^2 - 1} = \frac{24}{55}$, and $EX = \frac{1}{k^2 - 1} = \frac{1}{\left ( \frac{8}{3} \right )^2 - 1} = \frac{9}{55}$. Similarly, with $k=4$, we get $BY=\frac{4}{15}$ and $DB = \frac{1}{15}$. This reveals that $BC = BD + DE + ED = \frac{1}{15} + 1 + \frac{9}{55} = \frac{203}{165}$.

Now, note that $AX = AY = d$. We have $\frac{AC}{AB} = \frac{5}{4} \iff \frac{AX + XC}{AY+YB} = \frac{5}{4} \iff \frac{d+\frac{24}{55}}{d+\frac{4}{15}} = \frac{5}{4}$. This simplifies to $\frac{4}{3}+5d = \frac{96}{55} + 4d$, or alternatively $d = \frac{68}{165}$.

Finally, $\frac{BC}{AX} = \frac{\frac{203}{165}}{\frac{68}{165}} = \boxed{\frac{203}{68}}$.