HMMT 二月 2025 · 几何 · 第 10 题
HMMT February 2025 — Geometry — Problem 10
题目详情
- A plane P intersects a rectangular prism at a hexagon which has side lengths 45, 66, 63, 55, 54, and 77, in that order. Compute the distance from the center of the rectangular prism to P .
解析
- A plane P intersects a rectangular prism at a hexagon which has side lengths 45, 66, 63, 55, 54, and 77, in that order. Compute the distance from the center of the rectangular prism to P . Proposed by: Albert Wang, Karthik Venkata Vedula q 95 Answer: 24 Solution 1: Translate P so that it contains the center. The intersection of the translated plane with the rectangular prism is a centrally symmetric hexagon. Let its side lengths be a , b , c , a , b , and c , in that order. Then, for some t , t , and t , the side lengths of the hexagon before the translation were a b c ( a − t , b + t , c − t , a + t , b − t , c + t ) = (45 , 66 , 63 , 55 , 54 , 77) , a b c a b c from which it follows that t = 5, t = 6, and t = 7. a b c Now, a translation of a plane can be written in one of three equivalent forms: it can be viewed as a translation in the x direction by a distance d , a translation in the y direction by a distance d , or a x y translation in the z direction by a distance d (with coordinate axes chosen as shown below). z d x c d y O a a + t a b d z d x As shown above, we can express t , t , and t in terms of d , d , and d using the Pythagorean theorem, a b c x y z q q p 2 2 2 2 2 2 which yields t = d + d , t = d + d , and t = d + d . Hence, a b c y z z x x y 2 2 2 2 2 2 2 2 2 5 + 6 − 7 6 + 7 − 5 7 + 5 − 6 2 2 2 ( d , d , d ) = , , = (6 , 30 , 19) . x y z 2 2 2 We can draw a right pyramid with legs d , d , and d which has the center of the prism as a vertex x y z with opposite face on P . Then, the height of this pyramid, or the distance from the center to P , is s r r 1 1 95 = = . − 2 − 2 − 2 − 1 − 1 − 1 6 + 30 + 19 24 d + d + d x y z Solution 2: Let the vertices of the hexagon be ABCDEF , where AB = 45, BC = 66, etc. Note that AB ∥ DE , BC ∥ EF , and CD ∥ F A . Let O be the center of the prism, and let M , N , and P be the midpoints of AD , BE , and CF , respectively. E D M N F C P A B The key observation is that M N is the midline between AB and DE . Hence, plane OM N is the midplane between the faces of the prism containing sides AB and DE . Similarly, planes OM P and ON P are the other two midplanes of the prism. Thus, OM , ON , and OP are mutually orthogonal. Observe | AB − DE | | BC − EF | | CD − F A | M N = = 5 , N P = = 6 , and P M = = 7 , 2 2 2 p √ so by Heron’s formula, we can compute the area of M N P to be 9(9 − 5)(9 − 6)(9 − 7) = 6 6. Moreover, if x = OM , y = ON , and z = OP , then, 2 2 2 2 2 2 2 2 2 x + y = 5 , y + z = 6 , and z + x = 7 . √ √ √ Solving this system of equations gives x = 19, y = 6, and z = 30. Therefore, if d is the distance from O to plane M N P (i.e., the answer), the volume of tetrahedron OM N P can be written as √ √ √ √ 1 1 · 19 · 6 · 30 = · (6 6) · d, 6 3 so r √ 19 · 6 · 30 95 d = √ = . 24 2 · 6 6