Classic Geometry Problem

Assume that D:(0,0), A:(0,3q), B:(3p,3q) & C:(3p,0) -- a general rectangle where p. q are unknown. We find line BF to be: y=3qx/(2p) - 3q/2 and AC : x/(3p)+y/(3q)=1 which gives us Y to be: (9p/5, 6q/5). Similarly, determine X as (27p/13, 21q/13). We can now say that FY:YX:XB=(9/5 -1):(27/13 - 9/5)):(3 - 27/13) = 26:9:30

A coordinate free proof. By Tales theorem $\frac{|FY|}{|YB|} = \frac{|FC|}{|AB|} = \frac{2}{3}.$ that is $|FB| = \frac{5}{2}|FY|$ .

Let $G$ be the intersection point of lines $AX$ and $CD$ . Then again by Tales we have $\frac{|CG|}{|DC|} = \frac{|CG|}{|AB|}= \frac{|CE|}{|EB|} =\frac{1}{2}$ and $\frac{|FX|}{|XB|} = \frac{|FG|}{|AB|} = \frac{|FC| + |CG|}{|AB|} = \frac{2}{3} + \frac{1}{2} = \frac{7}{6},$ that is $|FB| = \frac{13}{7}|FX|$ . Hence we have that $\frac{|FY|}{|FX|} = \frac{|FB|}{|FX|}\,\frac{|FY|}{|FB|} = \frac{13}{7} \cdot \frac{2}{5} = \frac{26}{35},$ and $\frac{|YX|}{|FY|} = \frac{35 - 26}{35} = \frac{9}{26}.$ Now $\frac{3}{2}=\frac{|YB|}{|FY|} = \frac{|YX| + |XB|}{|FY|} = \frac{9}{26} + \frac{|XB|}{|FY|},$ hence $\frac{|XB|}{|FY|} = \frac{3}{2} - \frac{9}{26} = \frac{39 - 9}{26} = \frac{30}{26}$ so that $|XB|:|YX|:|FY| = 30:9:26$ . As $GCD\,(26,9,30)=1$ , we have that $a+b+c = 65$ .

Because $ABCD$ is an arbitrary rectangle, let's just assume that it's a square. From here, let's let the side length be $3$ . Now, we can graph this. We'll let Point $D$ lie on the origin, and so on and so forth, such that $B$ is located at $(3, 3)$ . Now, let's find the line $AE$ . We know that length $EC$ is equal to $1$ , so $E$ is the point $(3, 1)$ . Thus, the graph of the line is $y=\frac{-2}{3}x+3$ . Now let's find $FB$ . Similarly, we can label the point $F$ to be $(1, 0)$ . The graph of this line is $\frac{3}{2}x-\frac{3}{2}$ . To find point $X$ , let's set these equations equal. Solving, we find that $X$ is at the point $(\frac{27}{5}, \frac{33}{5})$ . We can likewise deduce that point $Y$ is $(\frac{9}{5}, \frac{6}{5})$ . Now, applying the distance formula, we find that the ratio is $26:9:30$ , and thus the desired answer is $26+9+30=\boxed{65}$ . And we're done! This is a really bad approach, the calculations are messy, it takes lots of time, etc., so try to find a cool way to tackle the problem. :D

a classic geometric sol n

So the required answer should be :: 26+9+30 = 65