## Task A-2.6.

Let $A B C D$ be a rectangle in which $|A B|=1$ and $|B C|=\sqrt{3}$. The inscribed circles of triangles $A B C$ and $A C D$ touch the segment $\overline{A C}$ at points $M$ and $N$. Determine $|M N|$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_ce71b02a7f068198719dg-14.jpg?height=605&width=414&top_left_y=1745&top_left_x=750)

From the Pythagorean theorem, $|A C|=\sqrt{|A B|^{2}+|B C|^{2}}=\sqrt{4}=2$.

Let $I$ be the center of the inscribed circle of triangle $A B C$, and let $E$ and $F$ be the points of tangency of the inscribed circle with sides $\overline{A B}$ and $\overline{B C}$, respectively. In quadrilateral $I E B F$, there are three right angles, and the sides $\overline{I E}$ and $\overline{I F}$ are radii of the inscribed circle. We conclude that this quadrilateral is a square, and the length of its side is the radius $r$ of the inscribed circle of triangle $A B C$.

The area of triangle $A B C$ can be calculated in two ways:

$$
P=\frac{|A B| \cdot|B C|}{2} \quad \text { and } \quad P=r \cdot s
$$

where $s$ is the semiperimeter of triangle $A B C$.

The semiperimeter is $s=\frac{1+2+\sqrt{3}}{2}=\frac{3+\sqrt{3}}{2}$. From the equality of the area of the triangle, we get

$$
\begin{aligned}
\frac{|A B| \cdot|B C|}{2} & =r \cdot s \\
\frac{\sqrt{3}}{2} & =r \cdot \frac{3+\sqrt{3}}{2} \\
r & =\frac{\sqrt{3}-1}{2}
\end{aligned}
$$

Both triangles $A M I$ and $A E I$ are right triangles with a common hypotenuse and one leg being the radius of the circle $r$. Therefore, they are congruent, and we have

$$
|A M|=|A E|=|A B|-|B E|=|A B|-r=1-\frac{\sqrt{3}-1}{2}=\frac{3-\sqrt{3}}{2}
$$

Triangles $C D A$ and $A B C$ are congruent, so by analogy, we get $|C N|=|A M|$. Finally,

$$
|M N|=|A C|-|A M|-|C N|=2-2 \cdot \frac{3-\sqrt{3}}{2}=\sqrt{3}-1
$$