PUMaC 2023 · 几何（B 组） · 第 5 题
PUMaC 2023 — Geometry (Division B) — Problem 5
题目详情
- Let △ ABC be a triangle with AB = 13, BC = 14, and CA = 15. Let D , E , and F be the midpoints of AB , BC , and CA respectively. Imagine cutting △ ABC out of paper and then folding △ AF D up along F D , folding △ BED up along DE , and folding △ CEF up along EF until A , B , and C coincide at a point G . The volume of the tetrahedron formed by vertices √ p q D , E , F , and G can be expressed as , where p , q , and r are positive integers, p and r are r relatively prime, and q is square-free. Find p + q + r .
解析
- Let △ ABC be a triangle with AB = 13, BC = 14, and CA = 15. Let D , E , and F be the midpoints of AB , BC , and CA respectively. Imagine cutting △ ABC out of paper and then folding △ AF D up along F D , folding △ BED up along DE , and folding △ CEF up along EF until A , B , and C coincide at a point G . The volume of the tetrahedron formed by vertices √ p q D , E , F , and G can be expressed as , where p , q , and r are positive integers, p and r are r relatively prime, and q is square-free. Find p + q + r . Proposed by Atharva Pathak Answer: 80 Let H be the foot of the perpendicular from A to DF and let H be the foot of the perpen- 1 2 dicular from E to DF . Note that a 13-14-15 triangle is a 5-12-13 triangle glued to a 9-12-15 triangle along the side of length 12. Because △ ADF and △ EF D are similar to △ ABC scaled 5 by a factor of 1 / 2, we get that AH = EH = 6, DH = F H = , and H H = 2. Let θ be 1 2 1 2 1 2 2 the dihedral angle between △ GDF and △ EDF in the tetrahedron. Because GE came from BE and CE in the original triangle, we have GE = 7. Now imagine projecting points G , H , 1 ′ H , and E onto a plane perpendicular to F D , such that G maps to G , H and H map to 2 1 2 ′ ′ H , and E maps to E . Since GE has a component of length H H = 2 perpendicular to the 1 2 √ √ ′ ′ ′ ′ ′ 2 2 plane, we get G E = 7 − 2 = 45. Applying the law of cosines to △ G H E with the ′ 3 angle θ at H gives cos θ = . So the height of the tetrahedron, which is the distance from 8 √ 3 55 ′ ′ ′ G to H E , is 6 sin θ = . Finally, the area of the base of the tetrahedron, i.e. △ DEF , is 4 √ √ 1 1 3 55 21 55 (7)(6) = 21, so the volume is (21) = , which gives a final answer of 80. 2 3 4 4