In the 1448th exercise, let the circles centered at $A$ and $B$ be denoted by $k_{1}$ and $k_{2}$, their radii by $r_{1}$ and $r_{2}$, respectively, and the radius of the circle $k$ that touches $k$ from the inside and $k_{1}$ and $k_{2}$ from the outside by $r_{3}$. What does the ratio $r_{3} / r_{1}$ tend to, if - with fixed $A, B$ - $C$ tends to $A$, that is, $r_{1}$ tends to 0?
In the solution of Exercise 1448, we saw that the distance $OC = t$ as a function of $(0 \leq t < 1)$, the radii in question, and their ratio are:

$$
r_{3}=\frac{1-t^{2}}{4}=\frac{(1-t)(1+t)}{4}, \quad r_{1}=AC=1-t, \quad \frac{r_{3}}{r_{1}}=\frac{1+t}{4}
$$

If now $C$ approaches $A$, then $CA = 1-t$ approaches 0, which means $t$ approaches 1, and thus the ratio in question approaches $1/2$.

We can also state our result as follows: if $C$ approaches $A$, then both $r_{3}$ and $r_{1}$ approach 0, but $r_{1}$ approaches 0 faster than $r_{3}$, so the ratio $r_{3} / r_{1}$, which was found to be $1/4$ at $t=0$, increases and can get arbitrarily close to $1/2$.[^0]


[^0]:    ${ }^{1}$ See the solution in this issue, on page 18.