A right-angled triangle with legs of 3 and 4 units has had its circumscribed circle drawn. What is the radius of the circle that touches the triangle's legs and the circumscribed circle from the inside?
I. solution. Let's draw the right-angled triangle $ABC$, where the legs $AC=3$ and $BC=4$ units long. We claim that the radius of the sought circle is 2 units. To prove this, it is clearly sufficient to show that the circle with a radius of 2 units, which touches the sides $AC$ and $BC$, also touches the circumcircle of the triangle $ABC$. The center of the circumcircle is the midpoint $F$ of side $AB$, and its radius is

$$
\frac{AB}{2}=\frac{\sqrt{AC^{2}+BC^{2}}}{2}=2.5
$$

units. The circle $k$ touches the circumcircle from the inside if and only if the distance between the centers of the two circles is $2.5-2=0.5$ units.

# 1984-02-067-1.eps 

## 1. figure

Since $k$ touches both $AC$ and $BC$, the center $O$ is 2 units away from both. Therefore, $O$ lies on the perpendicular bisector of $BC$, and $OD=2$. The point $F$, as the center of the circumcircle of triangle $ABC$, also lies on $OD$, and $FD=AC/2=1.5$, because $FD$ is the midline of triangle $ABC$. Therefore,

$$
OF=OD-FD=2-1.5=0.5
$$

and this is what we wanted to prove.

II. solution. We solve the problem in general. Let the radius of the sought circle $k$ be $r$, and its center be $O$. The point $O$ is $r$ units away from $AC$ and $BC$.

$$
\begin{gathered}
\text { 1984-02-067-2.eps } \\
\text { 2. figure }
\end{gathered}
$$

Drop perpendiculars from the midpoint $F$ of $AB$ to $AC$ and $BC$, and let the feet of these perpendiculars be $E$ and $D$ respectively (2. figure). These bisect the corresponding sides, so by the Pythagorean theorem for the right-angled triangle $OMF$,

$$
OF^{2}=OM^{2}+MF^{2}=(r-EC)^{2}+(CD-r)^{2}=\left(r-\frac{b}{2}\right)^{2}+\left(\frac{a}{2}-r\right)^{2}
$$

The circle $k$ touches the circumcircle from the inside if and only if $OF=AF-r=c/2-r$. From this, we get the following equation for $r$:

$$
\left(\frac{c}{2}-r\right)^{2}=\left(r-\frac{b}{2}\right)^{2}+\left(\frac{a}{2}-r\right)^{2}
$$

Using $a^{2}+b^{2}=c^{2}$, we get

$$
r(r-(a+b-c))=0
$$

Since $r \neq 0$, the radius of the sought circle is $r=a+b-c$. In our case, $a=3, b=4, c=5$, so $r=3+4-5=2$.