6. An equilateral triangle $A B C$ is given. On the side $B C$, find a point $P$ such that the incircle of triangle $A B P$ and the excircle opposite to side $P C$ of triangle $A P C$ are congruent.
SOLUTION. Let \( a \) denote the length of the side of the equilateral triangle \( ABC \). Further, let the lengths of the tangents from points \( A, B, C \), and \( P \) to both considered circles be denoted as shown in Fig. 3. Let the radii of both identical circles be denoted by \( r \). From the figure, it is evident that

\[
a + z = a - x + 2y, \quad \text{hence} \quad x + z = 2y.
\]

When expressing the length \( a \) of side \( BC \), we further obtain

![](https://cdn.mathpix.com/cropped/2024_04_17_888989f1419c479ab979g-4.jpg?height=401&width=398&top_left_y=1220&top_left_x=972)

\[
x + 2y + z = a.
\]

From equations (1) and (2), it immediately follows that

\[
x + z = 2y = \frac{a}{2}, \quad \text{thus} \quad y = \frac{a}{4}.
\]

We also see that the following pairs of relations hold:

\[
x = r \cot 30^\circ = r \sqrt{3} \quad \text{and} \quad z = r \cot 60^\circ = r \frac{\sqrt{3}}{3},
\]

from which it follows that

\[
x = 3z.
\]

Substituting (3) into (2) yields, after a simple rearrangement,

\[
\frac{a}{2} = x + z = 3z + z = 4z, \quad \text{hence} \quad z = \frac{a}{8}.
\]

Thus, we have

\[
|BP| = x + y = 3z + y = \frac{3}{8}a + \frac{1}{4}a = \frac{5}{8}a, \quad |CP| = a - |BP| = \frac{3}{8}a.
\]

From this, the construction of point \( P \) follows immediately.

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