Side lengths $a, b, c$ of right triangle satisfy $\frac{b}{c-a}+\frac{a}{c-b}=13/2$. What is $a:b:c$?

Question

In right triangle $ABC$, $a, b, c$ are the side lengths. c is the hypontenuse. If $\frac{b}{c-a}+\frac{a}{c-b}=\frac{13}{2}$ what is $a:b:c$?

Basically, this question is an algebra question with two equations:

$a^2+b^2=c^2$

$\frac{b}{c-a}+\frac{a}{c-b}=\frac{13}{2}$

Unfortunately, I'm pretty bad at algebra and I don't know how to start with this. Could someone please help me out?

Answer 1

Left $x=\frac ac$ and $y = \frac bc$. Then,

$$\frac y{1-x} + \frac x{1-y} = \frac {13}2$$ $$x^2+y^2=1$$

Rearrange to get

$$15(x+y) -13xy = 15$$ $$(x+y)^2-2xy=1$$

Solve for

$$x+y = \frac {17}{13}, \>\>\>\>\> xy = \frac {60}{169}$$

which yields,

$$x= \frac 5{13}, \>\>\>\> \>y = \frac {12}{13}$$

Thus, $a:b:c=5:12:13$.