<s>[INST] Area of a right triangle if $a/b$ is $1.05$ and the difference between the radii of the circumscribed and inscribed circles is $17$? What's the area of a right triangle if the quotient of its legs is $1.05$ and the difference between the radii of the inscribed and circumscribed circles is $17$?
I've been trying to solve this and I've got:
($R$ - radius of circumscribed circle, $r$ - radius of inscribed circle)
$1.$ $ \frac{a}{b}=1.05$
$2.$ $c^2=a^2+b^2$ 
$3.$ $a + b - 2r = c$ 
$4.$ $c-2r=34$
$5.$ $ab=(a+b+c)r$
Using the first four equations, I can substitute for one of the legs from $1.$ and for $r$ through $4.$ which leaves me with
$b(2.05)-2c=34$ 
$c=b\sqrt{1.05^2+1}$
However, solving this simply evades me, as I don't find myself getting rid of the square root which I don't know how to calculate.
I do know my equations give the right answer so I'm probably missing a simpler way to solve the system of equations.
Help much appreciated.
 [/INST] The inradius of a right triangle in terms of its sides is given by the relationship $$r \frac{a+b+c}{2} = rs =|\triangle ABC| = \frac{ab}{2},$$ hence $$r = \frac{ab}{a+b+c}.$$  The circumradius is trivially $$R = \frac{c}{2}.$$  Thus the given conditions may be summarized as $$17 = R - r = \frac{c}{2} - \frac{ab}{a+b+c} = \frac{(a+b)c + c^2 - 2ab}{2(a+b+c)} = \frac{(a+b)c + (a-b)^2}{2(a+b+c)}, \\ \frac{a}{b} = \frac{21}{20}, \\ a^2 + b^2 = c^2.$$  We note that because $21^2 + 20^2 = 29^2$, we have $$a : b  : c \equiv 21 : 20 : 29,$$ thus the computation is greatly simplified by letting $a = 21k$, $b = 20k$, $c = 29k$, to obtain from the first equation $$17(2)(70)k = (41)(29)k^2 + (21-20)^2 k^2,$$ or $$2380k = 1190k^2,$$ or $k = 2$.  Thus the desired triangle is $(a,b,c) = (42, 40, 58)$ and its area is $$|\triangle ABC| = 840.$$
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