6. On a plane, a segment of length 1 cm is given. Using a compass and a straightedge, construct a segment of length $\sqrt{\sqrt{3}+\sqrt{2}+1}$ cm on this plane.
Solution: On the line $\mathrm{AC}$, we mark off unit segments $\mathrm{AB}$ and $\mathrm{BC}$. We construct the perpendicular bisector $\mathrm{PB}$ of segment $\mathrm{AC}$ using a compass and straightedge. We mark off a unit segment $\mathrm{BK}$ on it and connect points $\mathrm{A}$ and $\mathrm{K}$. The length of segment $\mathrm{AK} = \sqrt{2}$. We mark off segment $\mathrm{PK} = \mathrm{AK}$ on line $\mathrm{PK}$ and connect points $\mathrm{A}$ and $\mathrm{P}$. The length of segment $\mathrm{BP} = \sqrt{3}$.

The following diagram shows the construction of segment $\mathrm{MB} = \sqrt{\sqrt{3} + \sqrt{2} + 1}$:

- We sequentially mark off segments $\mathrm{PA}$, $\mathrm{AK}$, $\mathrm{KB}$, and $\mathrm{BC}$ on a line with lengths $\sqrt{3}$, $\sqrt{2}$, $1$, and $1$, respectively.
- We construct the perpendicular bisector $\mathrm{HO}$ of segment $\mathrm{PC}$.
- We draw a circle with center at point $\mathrm{O}$ and radius $\mathrm{OP}$.
- We construct the perpendicular bisector $\mathrm{MB}$ of segment $\mathrm{KC}$.
- By the property of the right triangle $\mathrm{PMC}$, $\mathrm{MB} = \sqrt{\mathrm{PB} \cdot \mathrm{BC}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8ff2463fe8ad7c6cff62g-05.jpg?height=423&width=480&top_left_y=1353&top_left_x=434)
$C$

![](https://cdn.mathpix.com/cropped/2024_05_06_8ff2463fe8ad7c6cff62g-05.jpg?height=377&width=537&top_left_y=1402&top_left_x=1111)