## Task 3 - 130923

A convex isosceles trapezoid $A B C D (A B \| C D ; A D=B C ; A B>C D)$ should have the following properties:

It should be circumscribed around a circle with radius $r=12 \mathrm{~cm}$; the perimeter of the trapezoid should be $u=100 \mathrm{~cm}$.

Investigate whether such trapezoids exist and calculate the side lengths of each such trapezoid!
Since the trapezoid has an incircle, it is a tangential quadrilateral, so the sum of the lengths of each pair of its opposite sides is equal. Therefore, $|A B|+|C D|=|A D|+|B C|=\frac{u}{2}=50 \mathrm{~cm}$. Due to the isosceles property, it is already known that $|A D|=|B C|=25 \mathrm{~cm}$.

The radii of the incircle touching $A B$ and $C D$ are each perpendicular to the sides, so they are parallel to each other due to the parallelism of the two sides.

Since both radii pass through the center of the incircle, they complement each other to form a diameter that is perpendicular to $A B$ and $C D$, so these two parallels have a distance of $2 r=24$ $\mathrm{cm}$. In particular, they are therefore also shorter than $|A D|=|B C|=25 \mathrm{~cm}$.

Let $L_{C}$ and $L_{D}$ be the feet of the perpendiculars from $C$ and $D$ to $A B$. Then these perpendiculars $\left|C L_{C}\right|$ and $\left|D L_{D}\right|$ are each also $24 \mathrm{~cm}$, so they are of equal length. Thus, the right triangles $\triangle B L_{C} C$ and $\triangle A L_{D} D$ are congruent by the Ssw congruence theorem. Since the corresponding sides $C L_{C}$ and $D L_{D}$ are parallel and the corresponding sides $B L_{C}$ and $A L_{D}$ both lie on the line $A B$, there are only two possible positional relationships:

Either the third sides $A D$ and $B C$ of these two triangles are also parallel to each other, or they are reflected over the perpendicular bisector of the segment $A B$, pointing in different directions.

In the first of these two cases, in the quadrilateral $A B C D$, opposite sides are parallel, so it is a parallelogram. This implies $|A B|=|C D|$, which contradicts the problem statement.

In the second case, the feet of the perpendiculars $L_{C}$ and $L_{D}$ must both lie outside or both inside the segment $A B$, with the former leading to $|C D|>|A B|$, which contradicts the problem statement. Therefore, both feet of the perpendiculars lie inside the segment $A B$. Thus, $|A B|=\left|A L_{D}\right|+\left|L_{D} L_{C}\right|+$ $\left|L_{C} B\right|=\left|L_{D} L_{C}\right|+2\left|L_{C} B\right|$, the latter due to the congruence of the triangles $\triangle B L_{C} C$ and $\triangle A L_{D} D$.

In the quadrilateral $L_{D} L_{C} C D$, opposite sides are parallel, so it is a parallelogram and $\left|L_{D} L_{C}\right|=|C D|$ follows.

It remains to calculate $\left|L_{C} B\right|$. In the right triangle $\triangle B L_{C} C$, $\left|L_{C} C\right|^{2}+\left|B L_{C}\right|^{2}=|B C|^{2}$, so $\left|L_{C} B\right|=\sqrt{25^{2}-24^{2}} \mathrm{~cm}=\sqrt{49} \mathrm{~cm}=7 \mathrm{~cm}$.

Substituting this, we get $|A B|=|C D|+2 \cdot 7 \mathrm{~cm}$ and together with $|A B|+|C D|=50 \mathrm{~cm}$, we finally get $|A B|=32 \mathrm{~cm}$ and $|C D|=18 \mathrm{~cm}$. For such a trapezoid, its side lengths are $|A B|=32$ $\mathrm{cm},|B C|=25 \mathrm{~cm},|C D|=18 \mathrm{~cm}$ and $|A D|=25 \mathrm{~cm}$.