HMMT 十一月 2014 · 团队赛 · 第 10 题

HMMT November 2014 — Team Round — Problem 10

[ 8 ] Let ABCDEF be a convex hexagon with the following properties. (a) AC and AE trisect ∠ BAF . (b) BE ‖ CD and CF ‖ DE . (c) AB = 2 AC = 4 AE = 8 AF . Suppose that quadrilaterals ACDE and ADEF have area 2014 and 1400, respectively. Find the area of quadrilateral ABCD .

解析

[ 8 ] Let ABCDEF be a convex hexagon with the following properties. (a) AC and AE trisect ∠ BAF . (b) BE ‖ CD and CF ‖ DE . (c) AB = 2 AC = 4 AE = 8 AF . Suppose that quadrilaterals ACDE and ADEF have area 2014 and 1400, respectively. Find the area of quadrilateral ABCD . Answer: 7295 From conditions (a) and (c), we know that triangles AF E , AEC and ACB are similar to one another, each being twice as large as the preceding one in each dimension. Let AE ∩ F C = P and AC ∩ EB = Q . Then, since the quadrilaterals AF EC and AECB are similar to one another, we have AP : P E = AQ : QC . Therefore, P Q ‖ EC . Let P C ∩ QE = T . We know by condition (b) that BE ‖ CD and CF ‖ DE . Therefore, triangles P QT and ECD have their three sides parallel to one another, and so must be similar. From this we deduce that the three lines joining the corresponding vertices of the two triangles must meet at a point, i.e., that P E, T D, QC are concurrent. Since P E and QC intersect at A , the points A, T, D are collinear. Now, because T CDE is a parallelogram, T D bisects EC . Therefore, since A, T, D are collinear, AD also bisects EC . So the triangles ADE and ACD have equal area. Now, since the area of quadrilateral ACDE is 2014, the area of triangle ADE is 2014 / 2 = 1007. And since the area of quadrilateral ADEF is 1400, the area of triangle AF E is 1400 − 1007 = 393. Therefore, the area of quadrilateral ABCD is 16 · 393 + 1007 = 7295, as desired. Team Round