## Task 20/62

Given are $a$ and $b$ with $a>b$. It is required to construct $\frac{ab}{a-b}$.
The product $ab$ can be considered as the area $F$ of a rectangle $ABCD$ with sides $a$ and $b$. If we denote the desired quantity by $x$, then $\frac{ab}{a-b}=x$ or $x(a-b)=ab$.

Thus, $x$ can be seen as the side of a rectangle $A'B'C'D'$ with the area $F$ and the sides $x$ and $a-b$. The problem, therefore, represents a constructive area transformation: the rectangle $ABCD$ with sides $a$ and $b$ is to be transformed into an area-equivalent rectangle $A'B'C'D'$ with sides $x$ and $a-b$. There are several construction possibilities. The simplest one is as follows (figure):

1. Construct the rectangle $ABCD$ with $AB=a$ and

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-045.jpg?height=394&width=554&top_left_y=745&top_left_x=314)
$BC=b$.

2. Extend $BC$ beyond $C$ by $a-b$ to $E$.
3. Draw the lines through $A$ and $B$ as well as through $E$ and $D$ to intersect; let this intersection be $F$.
4. Construct perpendiculars to $AF$ at $F$ and to $CE$ at $E$; their intersection is $B'$.
5. Extend $AD$ to intersect $EB'$ at $A'$ and $CD$ to intersect $FB'$ at $C'$.

The rectangle $A'B'C'D'$ is the desired one.

It needs to be proven: 1. $ABCD = A'B'C'D'$, 2. $A'D = a-b$ or $C'D = a-b$.

1. $\triangle BEF \cong \triangle B'FE$ because $EF = EF$, $\measuredangle BFE = \measuredangle BE'F$, $\measuredangle BEF = \measuredangle B'FE$ (both are alternate interior angles of intersected parallels).

$\triangle AFD \cong \triangle C'DF$ because $DF = DF$, $\measuredangle AFB = \measuredangle C'DF$, $\measuredangle ADF = \measuredangle C'FD$ (both are alternate interior angles of intersected parallels).

$\triangle CDE \cong \triangle A'ED$ because $DE = DE$, $\measuredangle CDE = \measuredangle A'ED$, $\measuredangle CED = \measuredangle A'DE$ (both are alternate interior angles of intersected parallels).

From this, it follows:

$$
\square ABCD = \triangle BEF - \triangle AFD - \triangle CDE = \triangle B'FE - \triangle C'DF - \triangle A'ED = \square A'B'C'D'
$$

2. $A'D \parallel EC$, $A'E \parallel DC$, thus $A'D = EC = a-b$ (by construction).

## Solution by Klaus Müller:

Construction description:

The segment $AB = b$ is extended beyond $B$ by $a$ to point $C$. A perpendicular is constructed at $B$ to $AC$, intersecting the Thales circle over $AC$ at $D$. On $BC = a$, the segment $b = CE$ is marked from $C$.

The perpendicular bisector of $DE$ intersects the line through $A$ and $C$ at $F$. The arc around $F$ with radius $EF$ intersects the line through $A$ and $C$ at $G$. The segment $GB$ is the desired one.

Proof: $ab = DB^2$ (height theorem), $GB \cdot BE = DB^2$ (height theorem). $BE = a - b$ (by construction), thus $GB \cdot (a - b) = ab$, and therefore

$$
GB = \frac{ab}{a-b}
$$

## 2nd Solution by Klaus Müller:

Construction description:

Draw two rays from a point $S$ at any angle and mark the segment $SA = a$ on one of them and the segment $SB = a$ on the other.

On $AS$, mark the segment $AC = b$ from $A$. The parallel to $CB$ through $A$ intersects the ray $SB$ at $D$. The segment $BD$ is the desired one.

Proof: According to the first ray theorem, $BD : AC = BS : CS$, i.e.,

$$
BD = \frac{BS \cdot AC}{CS} = \frac{ab}{a-b}
$$

Solution by Hans-Jürgen Weiß:

The problem is solved if it is possible to construct a parallelogram with height $a-b$ that is area-equivalent to a rectangle with sides $a$ and $b$.

In the rectangle $ABCD$, let $AB = a$ and $BC = b (a > b)$. Draw a circular arc with radius $b$ around $B$, intersecting $AB$ at $E$. By construction, $AE = a - b$.

Now draw the Thales circle over $AB$ and a circular arc with radius $AE$ around $A$, intersecting the Thales circle at $F$. By construction, $AF = a - b$ and $AF \perp FB$. The line through $B$ and $F$ intersects the line through $C$ and $D$ at $C'$. The parallel through $A$ to $BC'$ intersects the line through $C$ and $D$ at $D'$.

The parallelogram $AB'C'D'$ shares the side $AB$ and the height $BC = AD$ with the rectangle, so both are area-equivalent. Furthermore, the parallelogram $AB'C'D'$ has the height $AF = a - b$ on the side $BC'$.

Therefore, $BC' = AD' = \frac{ab}{a-b}$ is the desired segment.