## 5. Pencil, Eraser, Protractor, and Compass

Take five items: a sheet of white paper, an eraser, a protractor, a compass, and a pencil. Try to accurately specify two segments whose sum of lengths equals the length of the pencil, and the square root of the product of their lengths equals the length of the eraser.
5. Place a pencil on paper. Mark its two extreme points $A$ and $B$ with a compass; connect these points with a segment using a protractor and a pencil.

Draw the perpendicular bisector of segment $A B$ (you, of course, know how to do this); thereby we determine the midpoint of $A B$. Then construct a semicircle with diameter $A B$. Using a protractor, draw a ray starting from point $A$, perpendicular to $A B$ and directed to the side of $A B$ where the semicircle is located.

![](https://cdn.mathpix.com/cropped/2024_05_21_fe999c0fe2ad81fc5164g-151.jpg?height=414&width=577&top_left_y=758&top_left_x=748)

Measure along this perpendicular from point $A$ the length of the eraser. Then apply the protractor to the eraser so that you can draw a perpendicular to the first perpendicular, parallel to $A B$ and passing through a point $G$ such that the length $A G$ equals the length of the eraser. This perpendicular will intersect the semicircle at two points; let $M$ be one of them.

Using a protractor, drop a perpendicular from $M$ to $A B$; let $H$ be the foot of this perpendicular. Then the segments $A H$ and $H B$ are the desired ones. Indeed, $A H + H B = A B =$ the length of the pencil. On the other hand, from the similarity of triangles $A M H$ and $M H B$ it follows that $A H : M H = M H : H B$, i.e., $A H \cdot H B = (M H)^2 =$ the square of the length of the eraser.