A tangent is drawn to the circle inscribed in an equilateral triangle with a side equal to $a$, intersecting two of its sides. Find the perimeter of the cut-off triangle.

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Apply the theorem of equality of tangent segments drawn from one point to a circle.

## Solution

Let the circle inscribed in an equilateral triangle with side length $a$ touch the sides $B C, A C$, and $A B$ at points $A_{1}, B_{1}$, and $C_{1}$ respectively, and let some line intersect the sides $A B$ and $A C$ at points $M$ and $N$ respectively and touch the circle at point $P$. Since $A C_{1}=A B_{1}$ and $A B=A C$, then $B C_{1}=C B_{1}$, and since $B A_{1}=B C_{1}$ and $C A_{1}=C B_{1}$, then $B A_{1}=C A_{1}$, i.e., $A_{1}$ is the midpoint of side $B C$. Similarly, we can prove that $C_{1}$ is the midpoint of $A B$ and $B_{1}$ is the midpoint of $A C$.

Since $M P=M C_{1}$ and $N P=N B_{1}$, then $M N=M P+N P=M C_{1}+N B_{1}$, therefore

$$
\begin{gathered}
A M+M N+A N=A M+\left(M C_{1}+N B_{1}\right)+A N=\left(A M+M C_{1}\right)+\left(N B_{1}+A N\right)= \\
=A C_{1}+A B_{1}=\frac{1}{2} a+\frac{1}{2} a=a .
\end{gathered}
$$

## Answer

a.