HMMT 二月 2018 · 几何 · 第 6 题
HMMT February 2018 — Geometry — Problem 6
题目详情
- Let ABC be an equilateral triangle of side length 1. For a real number 0 < x < 0 . 5, let A and A 1 2 be the points on side BC such that A B = A C = x , and let T = 4 AA A . Construct triangles 1 2 A 1 2 T = 4 BB B and T = 4 CC C similarly. B 1 2 C 1 2 There exist positive rational numbers b, c such that the region of points inside all three triangles T , T , T is a hexagon with area A B C √ 2 8 x − bx + c 3 · . (2 − x )( x + 1) 4 Find ( b, c ).
解析
- Let ABC be an equilateral triangle of side length 1. For a real number 0 < x < 0 . 5, let A and A 1 2 be the points on side BC such that A B = A C = x , and let T = 4 AA A . Construct triangles 1 2 A 1 2 T = 4 BB B and T = 4 CC C similarly. B 1 2 C 1 2 There exist positive rational numbers b, c such that the region of points inside all three triangles T , T , T is a hexagon with area A B C √ 2 8 x − bx + c 3 · . (2 − x )( x + 1) 4 Find ( b, c ). Proposed by: Kevin Sun Answer: (8 , 2) Solution 1: Notice that the given expression is defined and continuous not only on 0 < x < 0 . 5, but also on 0 ≤ x ≤ 0 . 5. Let f ( x ) be the function representing the area of the (possibly degenerate) hexagon for x ∈ [0 , 0 . 5]. Since f ( x ) is equal to the given expression over (0 , 0 . 5), we can conclude that f (0) and f (0 . 5) will also be equal to the expression when x = 0 and x = 0 . 5 respectively. (In other words, f ( x ) √ 3 is equal to the expression over [0 , 0 . 5].) In each of the cases, we can compute easily that f (0) = and 4 2 − b/ 2+ c c f (0 . 5) = 0, so by plugging them in, we get = 1 and = 0, which gives b = 8 and c = 2. 2 · 1 (3 / 2) · (3 / 2) Solution 2: Let P = AA ∩ CC , Q = AA ∩ BB , R = BB ∩ CC . These three points are the points on the hexagon 1 2 2 1 1 2 farthest away from A. For reasons of symmetry, the area of the hexagon (call it H for convenience) is: [ H ] = [ ABC ] − 3[ BP RQC ] . Also, [ BP C ] = [ BQC ] by symmetry, so: [ BP RQC ] = [ BP C ] + [ BQC ] − [ BRC ] [ BP RQC ] = 2[ BP C ] − [ BRC ] . From this, one can see that all we need to do is calculate the A -level of the points P and R in barycentric x x coordinates. Ultimately, the A -level of P is , and the A -level of R is . From this, straightforward x +1 2 − x calculation shows that: √ 2 8 x − 8 x + 2 3 [ H ] = · , (2 − x )( x + 1) 4 thus giving us the answer ( b, c ) = (8 , 2) .