HMMT 二月 2019 · 几何 · 第 4 题

HMMT February 2019 — Geometry — Problem 4

Convex hexagon ABCDEF is drawn in the plane such that ACDF and ABDE are parallelograms with area 168. AC and BD intersect at G . Given that the area of AGB is 10 more than the area of CGB , find the smallest possible area of hexagon ABCDEF .

解析

Convex hexagon ABCDEF is drawn in the plane such that ACDF and ABDE are parallelograms with area 168. AC and BD intersect at G . Given that the area of AGB is 10 more than the area of CGB , find the smallest possible area of hexagon ABCDEF . Proposed by: Andrew Lin Answer: 196 Since ACDF and ABDE have area 168, triangles ABD and ACD (which are each half a parallelogram) both have area 84. Thus, B and C are the same height away from AD , and since ABCDEF is convex, B and C are on the same side of AD . Thus, BC is parallel to AD , and ABCD is a trapezoid. In particular, we have that the area of ABG equals the area of CDG . Letting this quantity be x , we have that the area of BCG is x − 10, and the area of ADG is 84 − x . Then notice that [ ABG ] [ ADG ] AG x 84 − x 2 = = . This means that = . Simplifying, we have x − 47 x + 420 = 0; this [ CBG ] GC [ CDG ] x − 10 x has solutions x = 12 and x = 35. The area of ABCDEF is twice the area of trapezoid ABCD , or 2[ x + ( x − 10) + (84 − x ) + x ] = 4 x + 148; choosing x = 12, we get that the smallest possible area is 48 + 148 = 196.