## Task 3 - 200523

Fritz wants to draw four points $A, B, C, D$ in this order on a straight line. The following conditions must be met:

(1) The length of the segment $A D$ should be $15 \mathrm{~cm}$.

(2) The segment $B C$ should be $3 \mathrm{~cm}$ longer than the segment $A B$.

(3) The segment $C D$ should be twice as long as the segment $A C$.

Investigate whether these conditions can be met! If this is the case, determine all the length specifications for the segments $A B, B C$, and $C D$ that satisfy these conditions!
(I) If the conditions of the task are met and $A B=x \mathrm{~cm}$, then from (2) it follows that $B C=(x+3) \mathrm{cm}$, thus

$$
A C=A B+B C=x \mathrm{~cm}+(x+3) \mathrm{~cm}=(2 x+3) \mathrm{cm}
$$

Due to (3), it then follows that $C D=2 A C=2(2 x+3) \mathrm{cm}=(4 x+6) \mathrm{cm}$, and thus

$$
A D=A C+C D+(2 x+3) \mathrm{cm}+(4 x+6) \mathrm{cm}=(6 x+9) \mathrm{cm}
$$

Due to (1), it follows that $6 x+9=15$, so $6 x=6$ and thus $x=1$.

Therefore, the given conditions can only be met if $A B=1 \mathrm{~cm}$ and consequently (due to (2) and (3)) $B C=4 \mathrm{~cm}$ and $C D=10 \mathrm{~cm}$.

(II) That these length specifications indeed meet the given conditions is shown by the following check:

Since $A D=A B+B C+C D=1 \mathrm{~cm}+4 \mathrm{~cm}+10 \mathrm{~cm}=15 \mathrm{~cm}$, condition (1) is met.

The segment $B C$ is $3 \mathrm{~cm}$ longer than the segment $A B$, so condition (2) is also met.

Since $A C=A B+B C=5 \mathrm{~cm}$, the segment $C D$ is twice as long as the segment $A C$, and thus condition (3) is also met.