HMMT 二月 2022 · 几何 · 第 9 题
HMMT February 2022 — Geometry — Problem 9
题目详情
- Let A B C , A B C , and A B C be three triangles in the plane. For 1 ≤ i ≤ 3, let D , E , and 1 1 1 2 2 2 3 3 3 i i F be the midpoints of B C , A C , and A B , respectively. Furthermore, for 1 ≤ i ≤ 3 let G be the i i i i i i i i centroid of A B C . i i i Suppose that the areas of the triangles A A A , B B B , C C C , D D D , E E E , and F F F 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 are 2, 3, 4, 20, 21, and 2020, respectively. Compute the largest possible area of G G G . 1 2 3
解析
- Let A B C , A B C , and A B C be three triangles in the plane. For 1 ≤ i ≤ 3, let D , E , and 1 1 1 2 2 2 3 3 3 i i F be the midpoints of B C , A C , and A B , respectively. Furthermore, for 1 ≤ i ≤ 3 let G be the i i i i i i i i centroid of A B C . i i i Suppose that the areas of the triangles A A A , B B B , C C C , D D D , E E E , and F F F 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 are 2, 3, 4, 20, 21, and 2020, respectively. Compute the largest possible area of G G G . 1 2 3 Proposed by: Daniel Zhu Answer: 917 Solution: Let P ( x, y, z ) be the point with barycentric coordinates ( x, y, z ) in triangle A B C . Note i i i i that since this is linear in x , y , and z , the signed area of triangle P ( x, y, z ) P ( x, y, z ) P ( x, y, z ) is a 1 2 3 homogenous quadratic polynomial in x , y , and z ; call it f ( x, y, z ). We now claim that 1 1 1 1 1 1 4 f ( , , 0) + 4 f ( , 0 , ) + 4 f (0 , , ) − f (1 , 0 , 0) − f (0 , 1 , 0) − f (0 , 0 , 1) 1 1 1 2 2 2 2 2 2 f ( , , ) = . 3 3 3 9 2 2 2 This is easy to verify for f ∈ { x , y , z , xy, xz, yz } , after which the statement follows for general f by linearity. Then, assuming that we can arbitrarily choose the signs of the areas, the area is maximized at 4 · 2061 + 9 = 229 · 4 + 1 = 917 . 9 Now it remains to show that this best-case scenario is actually possible. The first step is to first show that these values from an actual f , i.e. that one can fit a homogenous quadratic polynomial through every six possible values for f at the six given points. One way to see this is to note that by choosing 2 2 2 the coefficients for x , y , and z , the values at the vertices of the triangle can be matched, while adding any of the xy , xz , and yz terms influences only one of the midpoints, so they can be matched as well. Now we show that this particular f can be realized by a choice of triangles. To do this, note that 1 1 by continuity there must exist x , y , and z with f ( x , y , z ) = 0, since f (1 , 0 , 0) and f ( , , 0) 0 0 0 0 0 0 2 2 are different signs, and introduce the new coordinates u = x − x and v = y − y ; then f can be 0 0 2 2 written as au + buv + cv + du + ev . Now, one can let P ( u, v ) = (0 , 0), P ( u, v ) = ( u, v ), and 1 2 P ( u, v ) = ( − cv − e, au + bv + d ). This can be shown to reproduce the desired f . 3 Finally, to address the condition that the original triangles must be nondegenerate, we can perturb each of the P by a constant, which doesn’t affect f as areas are translation-invariant. This concludes i the proof.