## Task 2 - 250732

Let $A B C$ be a right-angled triangle with $C$ as the vertex of the right angle and with $C A=4 \mathrm{~cm}$, $C B=20 \mathrm{~cm}$.

A natural number $x$ is required to satisfy the following conditions (1) and (2):

a) The segment $C B$ can be shortened by $x \mathrm{~cm}$; i.e., there is a point $B^{\prime}$ between $C$ and $B$ with $C B^{\prime}=C B-x \mathrm{~cm}$.

b) If at the same time the segment $C A$ is extended beyond $A$ by $x \mathrm{~cm}$ to a point $A^{\prime}$, then the area of the triangle $A^{\prime} B^{\prime} C$ is exactly $55 \%$ of the area of the triangle $A B C$.

Investigate whether there is exactly one natural number $x$ that satisfies conditions (1) and (2)! If this is the case, determine this number!
In the triangle $ABC$, because $CA \perp CB$, one of these two sides is simultaneously the height to the other. Therefore, the area of triangle $ABC$ is

$$
J=\frac{1}{2} \cdot CA \cdot CB=\frac{1}{2} \cdot 4 \cdot 20 \, \text{cm}^2=40 \, \text{cm}^2
$$

If $A' B' C$ is a triangle as described in (1) and (2), it is also right-angled at $C$ and thus has the area

$$
J'=\frac{1}{2} \cdot CA' \cdot CB'
$$

Furthermore, for $CA' = p \, \text{cm}$ and $CB' = q \, \text{cm}$, it holds that $p$ and $q$ are natural numbers with $p = 4 + x$ and $q = 20 - x$ (5).

From (3) and (4), it is clear that conditions (1) and (2) are satisfied if and only if, in addition to (5),

$$
J' = \frac{55}{100} J \quad \text{i.e.,} \quad p \cdot q = 44
$$

The only factorizations of 44 into a product of two natural numbers are $44 = 1 \cdot 44 = 2 \cdot 22 = 4 \cdot 11$.

From (5), it follows that $p + q = 24$ and $p > 4$, so the only possibility for (5) is $p = 22$, $q = 2$, and thus, from (5) again, $x = 18$.

Since this satisfies (5) and (6), it is proven:

There is exactly one natural number $x$ that satisfies (1) and (2), namely $x = 18$.