HMMT 十一月 2021 · 冲刺赛 · 第 15 题

HMMT November 2021 — Guts Round — Problem 15

[9] Tetrahedron ABCD has side lengths AB = 6 , BD = 6 2 , BC = 10 , AC = 8 , CD = 10 , and AD = 6 . √ a b The distance from vertex A to face BCD can be written as , where a, b, c are positive integers, b is c square-free, and gcd( a, c ) = 1. Find 100 a + 10 b + c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2021, November 13, 2021 — GUTS ROUND Organization Team Team ID#

解析

[9] Tetrahedron ABCD has side lengths AB = 6 , BD = 6 2 , BC = 10 , AC = 8 , CD = 10 , and √ a b AD = 6 . The distance from vertex A to face BCD can be written as , where a, b, c are positive c integers, b is square-free, and gcd( a, c ) = 1. Find 100 a + 10 b + c . Proposed by: Joseph Heerens Answer: 2851 Solution: First, we see that faces ABD, ABC, and ACD are all right triangles. Now, ABD can be visualized as the base, and it can be seen that side AC is then the height of the tetrahedron, as AC 2 6 should be perpendicular to both AB and AD. Therefore, the area of the base is = 18 and the 2 18 · 8 volume of the tetrahedron is = 48 . 3 √ √ √ 1 Now, let the height to BCD be h. The area of triangle BCD comes out to · 6 2 · 82 = 6 41 . This 2 means that the volume is √ √ √ 6 h 41 24 24 41 √ 48 = = 2 h 41 = ⇒ h = = . 3 41 41