HMMT 十一月 2014 · THM 赛 · 第 9 题

HMMT November 2014 — THM Round — Problem 9

In equilateral triangle ABC with side length 2, let the parabola with focus A and directrix BC intersect sides AB and AC at A and A , respectively. Similarly, let the parabola with focus B and directrix 1 2 CA intersect sides BC and BA at B and B , respectively. Finally, let the parabola with focus C and 1 2 directrix AB intersect sides CA and CB at C and C , respectively. 1 2 Find the perimeter of the triangle formed by lines A A , B B , C C . 1 2 1 2 1 2 k

解析

In equilateral triangle ABC with side length 2, let the parabola with focus A and directrix BC intersect sides AB and AC at A and A , respectively. Similarly, let the parabola with focus B and directrix 1 2 CA intersect sides BC and BA at B and B , respectively. Finally, let the parabola with focus C and 1 2 directrix AB intersect sides CA and CB at C and C , respectively. 1 2 Find the perimeter of the triangle formed by lines A A , B B , C C . 1 2 1 2 1 2 √ Answer: 66 − 36 3 Since everything is equilateral it’s easy to find the side length of the wanted triangle. By symmetry, it’s just AA + 2 A B = 3 AA − AB . Using the definition of a parabola, 1 1 2 1 √ √ 3 AA = A B so some calculation gives a side length of 2(11 − 6 3), thus the perimeter claimed. 1 1 2 1 2 2 This can be seen rigorously by showing that five points in the plane determine at most one quadratic equation ax + by + cxy + dx + ey + f = 0, up to scaling. Theme Round k