2
$\begingroup$

Let $T$ be a circle with diameter $AB$. Let $P$ be a point inside the circle such that P lies on the line $AB$. Consider the circles wit diameters $PA=6$ and $PB=4$. A fourth circle $r$ is drawn such that it is tangent to the previous three circles. Prove that the radius of $r$ exceeds $3/2$.

I assumed radius of $r$ to be $k$. Then as the line joining the centers of two circles touching each other at one point passes through the point of contact, we obtain a triangle with sides $3+a, 5$ and $2+a$ (if I havent gone wrong anywhere). But after that I dont know how to proceed. Please help.

$\endgroup$
3
$\begingroup$

Both the Soddy-Gosset Theorem and Descartes' Theorem say $$ \left(\frac12+\frac13-\frac15+\frac1r\right)^2=2\left(\frac1{2^2}+\frac1{3^2}+\frac1{5^2}+\frac1{r^2}\right) $$ which means that $r=\frac{30}{19}\gt\frac32$.

$\endgroup$
1
$\begingroup$

Hint. Let $A=(-5,0)$ and let $B=(5,0)$. Then $P=(1,0)$ and if $(x,y)$ is the center of the fourth circle and $k$ is its radius, it follows that $$\begin{cases} (x+2)^2+y^2=(k+3)^2\\(x-3)^2+y^2=(k+2)^2\\x^2+y^2=(5-k)^2 \end{cases}.$$ Now you should be able to find the precise value of $k$.

$\endgroup$
  • $\begingroup$ Ah, coordinate bashing (if I am not wrong) $\endgroup$ – Yellow Jan 10 at 8:08
  • $\begingroup$ See my edit. Note that by solving the system we get $k=30/19$. $\endgroup$ – Robert Z Jan 10 at 8:24
  • $\begingroup$ Oh wait, what do you mean by $k$ is its center? $\endgroup$ – Yellow Jan 10 at 8:37
  • $\begingroup$ Sorry $k$ is the radius. The same notation you used in your question. $\endgroup$ – Robert Z Jan 10 at 8:40

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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