In a right-angled triangle, the radius of the circumscribed circle is $14.5 \mathrm{~cm}$ and the radius of the inscribed circle is $6 \mathrm{~cm}$.

Determine the perimeter of this triangle.

(L. Hozová)
The center of the circumcircle of a right-angled triangle is at the midpoint of its hypotenuse; this point is marked as $O$ in the diagram. The center of the incircle is the common intersection of the angle bisectors of the triangle; this point is marked as $V$, and the points of tangency of the circle with the individual sides of the triangle are marked as $X, Y$, and $Z$:

![](https://cdn.mathpix.com/cropped/2024_04_17_d6446605e32054ca6a34g-4.jpg?height=443&width=822&top_left_y=1960&top_left_x=617)

The radius of the circumcircle is $14.5 \, \text{cm}$, and the hypotenuse $KM$ is the diameter of this circle, so the length of $KM$ is $29 \, \text{cm}$.

The radius of the incircle is $6 \, \text{cm}$, and this is also the length of the segments $VX, VY$, and $VZ$. The congruent segments $VX$ and $VZ$ are adjacent sides of the rectangle $VXLZ$, making this rectangle a square, and the lengths of the segments $LX$ and $LZ$ are also $6 \, \text{cm}$.

Triangles $KYV$ and $KZV$ are both right-angled, have congruent angles at vertex $K$, and share the side $KV$. Therefore, these triangles are congruent, so the segments $KY$ and $KZ$ are congruent; let's denote their length as $a$. For the same reason, the segments $MX$ and $MY$ are also congruent; let's denote their length as $b$.

Thus, we have $a + b = |KM| = 29 \, \text{cm}$, and the perimeter of triangle $KLM$ can be expressed as

$$
2a + 2b + 2 \cdot 6 = 2(a + b) + 12 = 70 \, \text{cm}.
$$

Scoring: 1 point for the length of the hypotenuse; 3 points for expressing the perimeter; 2 points for the quality of the commentary.

Note. If we denote the radius of the circumcircle as $R$ and the length of the hypotenuse as $|KM| = l$, then the first observation in the given solution can be generally written as $l = 2R$, i.e.,

$$
R = \frac{l}{2}
$$

If we further denote the radius of the incircle as $r$ and the lengths of the legs as $|LM| = k$ and $|KL| = m$, then the subsequent conclusions from the previous solution are $m = r + a$, $k = r + b$, and $l = a + b$:

![](https://cdn.mathpix.com/cropped/2024_04_17_d6446605e32054ca6a34g-5.jpg?height=434&width=826&top_left_y=1371&top_left_x=615)

From this, it follows that $k + m - l = 2r$, i.e.,

$$
r = \frac{k + m - l}{2}
$$

and the perimeter of the triangle can be generally expressed as

$$
k + m + l = 2r + 4R.
$$

Formulas (1) and (2) can be considered known and appear in some tables. A solution based on them (3) should therefore be considered correct.