## Intuition The problem revolves around maximizing the inner radius \( r2 \) of a medal, given constraints on the outer radius \( r1 \), and the densities \( p1 \) and \( p2 \) of the materials used. The key insight is to understand the relationship between the masses of the ring and the disk, and how they relate to the given constants \( A \) and \( B \). The mass of the ring is proportional to the difference in areas between the outer circle and the inner circle, while the mass of the disk is proportional to the area of the inner circle. The ratio of these masses must equal \( \frac{A}{B} \). ## Approach 1. **Understand the Mass Relationship:** - The mass of the ring \( m_{\text{out}} \) is given by: \[ m_{\text{out}} = \pi \cdot (r1^2 - r2^2) \cdot p1 \] - The mass of the disk \( m_{\text{in}} \) is given by: \[ m_{\text{in}} = \pi \cdot r2^2 \cdot p2 \] - The ratio of these masses must satisfy: \[ \frac{m_{\text{out}}}{m_{\text{in}}} = \frac{A}{B} \] 2. **Derive the Formula for \( r2 \):** - From the mass ratio equation, we get: \[ \frac{(r1^2 - r2^2) \cdot p1}{r2^2 \cdot p2} = \frac{A}{B} \] - Rearrange to solve for \( r2^2 \): \[ r2^2 = \frac{B \cdot r1^2 \cdot p2}{A \cdot p2 + B \cdot p1} \] 3. **Maximize \( r2 \):** - Iterate over all possible values of \( r1 \), \( p1 \), and \( p2 \). - For each combination, compute \( r2^2 \) using the derived formula. - Track the maximum value of \( r2 \). 4. **Output the Result:** - Print the maximum \( r2 \) found, ensuring precision to meet the problem's requirements. ## Complexity - **Time:** \( O(n \times m \times k) \) where \( n \), \( m \), and \( k \) are the sizes of the lists for \( r1 \), \( p1 \), and \( p2 \) respectively. This is because we evaluate each combination of these parameters. - **Space:** \( O(1) \) as we only need a constant amount of additional space to store intermediate results. ## Solution (C++) ```cpp #include using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n, m, k; cin >> n; vector r1_options(n); for (int i = 0; i < n; ++i) { cin >> r1_options[i]; } cin >> m; vector p1_options(m); for (int i = 0; i < m; ++i) { cin >> p1_options[i]; } cin >> k; vector p2_options(k); for (int i = 0; i < k; ++i) { cin >> p2_options[i]; } int A, B; cin >> A >> B; double max_r2 = 0.0; for (int r1 : r1_options) { for (int p1 : p1_options) { for (int p2 : p2_options) { double r2_squared = (B * r1 * r1 * p2) / (A * p2 + B * p1); double r2 = sqrt(r2_squared); max_r2 = max(max_r2, r2); } } } cout << fixed << setprecision(9) << max_r2 << '\n'; return 0; } ``` This solution efficiently computes the maximum possible \( r2 \) by evaluating all combinations of the given parameters and using the derived formula to calculate \( r2 \). The use of precision formatting ensures the output meets the problem's requirements for accuracy.

Olympic Medal

Problem Details