MHB Circumcircles Find a formula relating R,R_1,R_2.

[mrtwhs]

Let $$\triangle ABC$$ be a right triangle with right angle at $$C$$. Suppose this right triangle has legs $$BC=a$$, $$AC=b$$, hypotenuse $$AB=C$$, and circumradius $$R$$. Two copies of this triangle can be joined to form an isosceles triangle in two ways. With $$a$$ as a common side, you can form an isosceles triangle with sides $$c,c,2b$$ and circumradius $$R_1$$. With $$b$$ as a common side, you can form an isosceles triangle with sides $$c,c,2a$$ and circumradius $$R_2$$. Find a formula relating $$R,R_1,R_2$$.

 

[Olinguito]

My attempt:

Spoiler

For the first circle, we clearly have
$$R=\frac c2$$
as the circumcentre is the midpoint of the hypotenuse $\mathrm{AB}$.

For the second circle, let $\mathrm D$ be the point on $\mathrm{BC}$ extended so that $\angle\,\mathrm{DAB}$ is a right angle. Then $\triangle\mathrm{DCA}$ is similar to $\triangle\mathrm{ACB}$ and so $|\mathrm{DC}|=\dfrac{b^2}a$. The circumcentre is the midpoint of $\mathrm{DB}$ and so
$$R_1=\frac12\left(\frac{b^2}a+a\right)=\frac{c^2}{2a}.$$

By symmetry,
$$R_2=\frac{c^2}{2b}.$$

Hence:
$$c^2=a^2+b^2=\frac{c^4}{4R_1^2}+\frac{c^4}{4R_2^2}$$
$\implies\ \dfrac1{c^2}=\dfrac1{4R_1^2}+\dfrac1{4R_2^2}$

$\implies\ \boxed{\dfrac1{R^2}\ =\ \dfrac1{R_1^2}+\dfrac1{R_2^2}}$.

 

[mrtwhs]

Spoiler

$\boxed{\dfrac1{R^2}\ =\ \dfrac1{R_1^2}+\dfrac1{R_2^2}}$.

Nice! Pretty much the same way I solved it.