The figure below shows a segment $AB$, its midpoint $C$, and the semicircles with diameters $AB$ and $AC$. A circle with center $P$ is tangent to both semicircles and also to the segment $AB$. Given $AB=8 \text{ cm}$, and $O$, the midpoint of $AC$, the questions are:

![](https://cdn.mathpix.com/cropped/2024_05_01_42ab6d7ce4e674d231dcg-11.jpg?height=463&width=830&top_left_y=567&top_left_x=693)

a) What is the measure of the perimeter of triangle $OCP$?

b) What is the measure of the radius of the circle with center $P$?
Solution

a) Let $x$ be the radius of the circle centered at $P$. We draw $OP$, which passes through the point of tangency $D$; $CP$, which passes through the point of tangency $F$; and $PE$, perpendicular to $AB$ (and a radius of the circle highlighted in the figure). We have $CP=4-x$ and $OP=2+x$. The perimeter of triangle $OCP$ is

$$
\begin{aligned}
2 p & =O P+C P+C O \\
& =(2+x)+(4-x)+2 \\
& =8 \text{ cm}
\end{aligned}
$$

b) Now, we will calculate the area of $\triangle OCP$ in two ways: using Heron's formula, which uses the semi-perimeter $p=4 \text{ cm}$, and as half the product of the base by the height.

![](https://cdn.mathpix.com/cropped/2024_05_01_42ab6d7ce4e674d231dcg-11.jpg?height=468&width=838&top_left_y=2005&top_left_x=710)

Thus,

$$
\sqrt{4 \cdot(4-2)(4-(2+x))(4-(4-x))}=\frac{2 \cdot x}{2}
$$

Solving the equation above, we get $x=\frac{16}{9} \text{ cm}$.

Note: A triangle with perimeter $2 p$ and side lengths $a, b$, and $c$, by Heron's Formula, has an area given by

$$
A=\sqrt{p(p-a)(p-b)(p-c)}
$$