\[
\left.\begin{array}{l}
\text { Tangent Circles } \\
\text { [ Area of a Circle, Sector, and Segment ]}
\end{array}\right]
\]

On a line, three distinct points $L, M$, and $N$ are taken ($M$ is between $L$ and $N$, $L N \neq M N$). On the segments $L M, M N$, and $L N$ as diameters, semicircles are constructed, with midpoints at points $A, B$, and $C$ respectively. Point $C$ lies on one side, and points $A$ and $B$ on the other side of the line $L N$. Find the ratio of the area of the figure bounded by these three semicircles to the area of triangle $A B C$.
Let $E$ and $F$ be the projections of the centers of the semicircles with diameters $L M=2 r$ and $M N=2 R$ onto the tangent to the semicircle with diameter $L N$, drawn through point $C$. Let $P$ be the projection of point $B$ onto the line $A E$. Then the area of the triangle is equal to the difference between the area of the rectangle $P E F B$ and the areas of three right triangles.

## Solution

Let $O_{1}, O_{2}$, and $O$ be the centers of the circles with diameters $L M, M N$, and $L N$ respectively, with radii $r$ and $R$ for the first and second circles respectively. Then the radius of the circle with center $O$ is $r+R$. It is known that $r \neq R$. Assume that $r < R$.

The figure described in the problem consists of semicircles with radii $r, R$, and $R+r$. If $S$ is its area, then

$$
S=\frac{1}{2} \pi r^{2}+\frac{1}{2} \pi R^{2}+\frac{1}{2} \pi(r+R)^{2}=\pi\left(R^{2}+r R+r^{2}\right)
$$

Let $E$ and $F$ be the projections of points $O_{1}$ and $O_{2}$ respectively onto the tangent to the semicircle with center $O$, drawn through point $C$. Since points $A$ and $B$ are the midpoints of the corresponding semicircles, points $A, O_{1}$, and $E$ lie on the same line, and points $B, O_{2}$, and $F$ also lie on the same line.

Let $P$ be the projection of point $B$ onto the line $A E$. Since $O_{1} A=r < R=O_{2} B$, point $A$ lies between points $O_{1}$ and $P$, and $O_{1} P=R-r$.

Note that the area of triangle $A B C$ is equal to the area of the rectangle $P E F B$ minus the areas of three right triangles $O_{1} P_{2}, B E O_{1}$, and $B F O_{2}$. Since

$$
C E=O O_{1}=R, C F=O O_{2}=r, E F=r+R, B F=2 R+r, A E=2 r+R
$$

then

$$
\begin{aligned}
& S_{\mathrm{PEFB}}=E F \cdot F B=(r+R)(2 R+r), S_{\triangle \mathrm{APB}}=\frac{1}{2} \cdot P B \cdot P A=\frac{1}{2}(R+r)(R-r) \\
& S_{\triangle \mathrm{AEC}}=\frac{1}{2} \cdot C E \cdot E A=\frac{1}{2} R(2 r+R), S_{\Delta \mathrm{BFC}}=\frac{1}{2} \cdot C F \cdot F B=\frac{1}{2} r(2 R+r)
\end{aligned}
$$

Therefore,

$$
S_{\triangle \mathrm{ABC}}=(r+R)(2 R+r)-\frac{1}{2}((R+r)(R-r)+R(2 r+R)+r(2 R+r))=R^{2}+r R+r^{2}
$$

Thus,

$$
\frac{S}{S_{\triangle A B C}}=\frac{\pi\left(R^{2}+R r+r^{2}\right)}{R^{2}+r R+r^{2}}=\pi
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f7e07b07084714df1adcg-41.jpg?height=783&width=820&top_left_y=0&top_left_x=619)

## Answer

$\pi$