In Kocourkov, they planned to build a decorative bridge over the river, whose arch was to be part of a circle. The profile of the arch, together with the roadway, forms a circular segment. However, in the original design, the height of the bridge arch was too large. They therefore built a bridge whose arch height was three times smaller, which, however, doubled the radius of the corresponding circle. In what ratio was the height of the bridge arch to the radius of the corresponding circle in the design and what was the ratio in the built bridge?

![](https://cdn.mathpix.com/cropped/2024_04_17_d40be33f9c82eb5d598cg-5.jpg?height=277&width=400&top_left_y=798&top_left_x=862)

(M. Petrová)
It is necessary to realize that the length of the bridge $l$ is the same in both cases. Let $r_{1}$ and $v_{1}$ be the radius of the corresponding circle and the height of the arch of the proposed bridge, and similarly, $r_{2}$ and $v_{2}$ denote the corresponding quantities for the built bridge.
![](https://cdn.mathpix.com/cropped/2024_04_17_d40be33f9c82eb5d598cg-5.jpg?height=394&width=1226&top_left_y=1402&top_left_x=446)

The center of the arch circle and the endpoints of the bridge roadway form an isosceles triangle, with the base being the bridge roadway (the bridge roadway is a chord of the corresponding circle). Then, for the length of the bridge $l$, according to the Pythagorean theorem, we have

$$
\begin{aligned}
& \left(\frac{l}{2}\right)^{2}=r_{1}^{2}-\left(r_{1}-v_{1}\right)^{2}=2 r_{1} v_{1}-v_{1}^{2}, \\
& \left(\frac{l}{2}\right)^{2}=r_{2}^{2}-\left(r_{2}-v_{2}\right)^{2}=2 r_{2} v_{2}-v_{2}^{2} .
\end{aligned}
$$

We compare equations (1) and (2):

$$
2 r_{1} v_{1}-v_{1}^{2}=2 r_{2} v_{2}-v_{2}^{2}
$$

From the problem statement, we also know that

$$
\begin{aligned}
& v_{1}=3 v_{2}, \\
& r_{2}=2 r_{1} .
\end{aligned}
$$

Substitute into equation (3) and further simplify:

$$
\begin{aligned}
2 r_{1}\left(3 v_{2}\right)-\left(3 v_{2}\right)^{2} & =2\left(2 r_{1}\right) v_{2}-v_{2}^{2} \\
6 r_{1} v_{2}-9 v_{2}^{2} & =4 r_{1} v_{2}-v_{2}^{2} \\
2 r_{1} v_{2} & =8 v_{2}^{2} \\
r_{1} v_{2} & =4 v_{2}^{2}
\end{aligned}
$$

Since $v_{2}$ is a positive number, we can divide the equation by it:

$$
r_{1}=4 v_{2} \text {. }
$$

Using relationship (5):

$$
r_{2}=2 r_{1}=2 \cdot 4 v_{2}=8 v_{2} \text {. }
$$

Now we determine the desired ratios:

- the ratio for the design is $v_{1}: r_{1}=\left(3 v_{2}\right):\left(4 v_{2}\right)=3: 4$,
- the ratio for the built bridge is $v_{2}: r_{2}=v_{2}:\left(8 v_{2}\right)=1: 8$.

Evaluation. 1 point for correctly using the Pythagorean theorem; 1 point for comparing the lengths of the bridge; 1 point for relationships (4) and (5); 1 point for deriving relationship (6) or an analogous relationship; 1 point for each final ratio.