5. The edges of the tetrahedron $ABCD$ have lengths 7, 13, 18, 27, 36, and 41 (in some order). If $AB$ has a length of 41, determine the length of the edge $CD$.

## Fourth grade - B category
5. Each side of the tetrahedron is a triangle, and the lengths of the sides of each of these triangles must satisfy the triangle inequality. This means \( AC + CB > AB = 41 \) and \( AD + DB > AB = 41 \). Therefore, we get \( AC + CB + AD + DB > 82 \). This means that both edges 27 and 36 must be among \( AC, CB, AD \) or \( DB \) (indeed, if this were not satisfied, then the sum \( AC + CB + AD + DB \) would be \( 7 + 13 + 18 + 27 \) or \( 7 + 13 + 18 + 36 \), i.e., 65 or 74, which is not greater than 82). Moreover, both of these edges cannot simultaneously be sides of \( \triangle ABC \) and \( \triangle ABD \), because then in the other triangle, the side \( AB = 41 \) would have to be less than the sum of the other two, and that sum would be at most \( 13 + 18 = 31 \), a contradiction. Therefore, in \( \triangle ABC \), one side has a length of 27, and let this be, without loss of generality, \( AC \), and then in \( \triangle ABD \), one side has a length of 36. In \( \triangle ABC \), it must hold that \( BC > AB - AC = 41 - 27 = 14 \), so the only remaining possibility is \( BC = 18 \). Now, if \( BD = 36 \), from \( \triangle BCD \) we would have \( CD > BD - BC = 36 - 18 = 18 \), which is impossible because there are no more "free" edges longer than 18. Therefore, in \( \triangle ABD \), the side \( BD \) cannot have a length of 36, so it must be \( AD = 36 \). Finally, from \( \triangle ACD \) we get \( CD > AD - AC = 36 - 27 = 9 \), so the only remaining possibility is
\( CD = 13 \)

## Fourth Grade - B Category