HMMT 二月 2016 · 冲刺赛 · 第 27 题

HMMT February 2016 — Guts Round — Problem 27

[ 14 ] Find the smallest possible area of an ellipse passing through (2 , 0), (0 , 3), (0 , 7), and (6 , 0).

解析

[ 14 ] Find the smallest possible area of an ellipse passing through (2 , 0), (0 , 3), (0 , 7), and (6 , 0). Proposed by: Calvin Deng √ 56 π 3 Answer: 9 Let Γ be an ellipse passing through A = (2 , 0) , B = (0 , 3) , C = (0 , 7) , D = (6 , 0), and let P = (0 , 0) be Area of Γ the intersection of AD and BC . is unchanged under an affine transformation, so we just Area of ABCD P A 1 P B 3 have to minimize this quantity over situations where Γ is a circle and = and = . In fact, P D 3 BC 7 √ √ we may assume that P A = 7 , P B = 3 , P C = 7 , P D = 3 7. If ∠ P = θ , then we can compute lengths to get √ 2 Area of Γ 32 − 20 7 cos θ + 21 cos θ r = = π √ 3 Area of ABCD 9 7 · sin θ Let x = cos θ . Then if we treat r as a function of x , √ ′ r 3 x 42 x − 20 7 0 = = + √ 2 2 r 1 − x 32 − 20 x 7 + 21 x √ √ √ 3 which means that 21 x − 40 x 7 + 138 x − 20 7 = 0. Letting y = x 7 gives 3 2 2 0 = 3 y − 40 y + 138 y − 140 = ( y − 2)(3 y − 34 y + 70) √ The other quadratic has roots that are greater than 7, which means that the minimum ratio is y 2 Area of Γ √ √ attained when cos θ = x = = . Plugging that back in gives that the optimum is Area of ABCD 7 7 √ √ 28 π 3 56 π 3 , so putting this back into the original configuration gives Area of Γ ≥ . If you want to 81 9 8 7 check on Geogebra, this minimum occurs when the center of Γ is ( , ). 3 3