PUMaC 2013 · 几何（A 组） · 第 8 题
PUMaC 2013 — Geometry (Division A) — Problem 8
题目详情
- [ 8 ] Three chords of a sphere, each having length 5 , 6 , 7, intersect at a single point inside the sphere and are pairwise perpendicular. For R the minimum possible radius of the sphere, find 2 R . 1
解析
- [ 8 ] Three chords of a sphere, each having length 5 , 6 , 7, intersect at a single point inside the sphere and are pairwise perpendicular. For R the minimum possible radius of the sphere, find 2 R . Solution Let X be the intersection point, A B , A , B , A B be three chords of length 1 1 2 2 3 3 ( l , l , l ) = (5 , 6 , 7) respectively, and M , M , M be their midpoints respectively. The notion 1 2 3 1 2 3 2 of ”power of a point” generalizes naturally to sphere, as any two chords intersecting at X should lie on a same plane and power theorem for the circle section can be applied. Thus we can write p = A X · B X = A X · B X = A X · B X. 1 1 2 2 3 3 2 2 It also holds that p = R OX since you can still calculate the power using the diameter 2 2 going through X . Also we have A X · B X = A M XM , thus the length t = XM is i i i i i i i given by 2 2 l l 2 2 i 2 i p = R OX = t , t = p. i i 4 4 Meanwhile note that that M O is prependicular to A B , thus XM is projection of XO onto i i i i A B . As three chords are pairwise perpendicular, the Pythagorean theorem gives i i 2 2 2 l + l + l 2 2 2 2 1 2 3 OX = t + t + t = 3 p 1 2 3 4 so 2 2 2 l + l + l 1 2 3 2 2 R = OX + p = 2 p. 4 2 2 2 Now p = l / 4 t is at most l / 4 = 25 / 4, and this bound gives 1 1 1 2 2 2 l + l + l 25 25 + 36 + 49 50 2 1 2 3 R 2 = = 15 . 4 4 4 Notice that this bound is indeed attainable. For any values of t , t , t and p > 0 satisfying 1 2 3 above equations, the relative positions of A s and B s are all fixed. And if we define the point i i ! ! ! ! 2 2 O to be XO = XM + XM + XM , then it can be proven that OA = OB = const using 1 2 2 i i the Pythagorean theorem. 3