Two circles $C_1$ and $C_2$ with radius $r_1$ and $r_2$ touching each other externally and both touching circle $C_3$ with radius $r_1+r_2$ internally. If another circle with radius $r_3$ touches all three circles such that $r_1, r_2, r_3$ are in $A.P.$ and $r_1\gt r_2 \gt r_3$ find $$\frac{r_1}{r_2}$$

I tried applying the Descartes' theorem but the equation so formed seemed to be nearly unsolvable. Any new method or improvement in the method of Descartes theorem is appreciated.

  • $\begingroup$ What is $A$?$ $ $\endgroup$ – Bernard Dec 20 '17 at 16:32
  • $\begingroup$ It is not just A. It is A.P. i.e arithmetic progression $\endgroup$ – Rohan Shinde Dec 20 '17 at 16:34

Say $r_1 = k r_2$, so $r_3 = (2-k) r_2$. Observe first that the centres of $C_1$, $C_2$ and $C_3$ lie on a straight line, by considering the radii of each, so our diagram looks like this, where $D$ is the centre of $C_3$ enter image description here From here, notice that $AD = r_1 + r_2 - r_3 = (2k-1)r_2$, and $CD= r_1 + r_2 - r_1 = r_2$. In particular, we know every length inside the triangle in terms of $r_2 $and $k$. Apply Stewart's Theorem and cancel $r_2^3$ to give $(k+1)k + (2k-1)^2 (k+1) = 4k + (3-k)^2$, which simplifies to $k^3 = 2$, so $k = \sqrt[3]{2} = \frac{r_1}{r_2}$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.