Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Prove it Taiwan

There are 2020 points {Ai=(xi,yi):i=1,,2020}\{A_i = (x_i, y_i) : i = 1, \dots, 2020\} on the plane, satisfying
0=x1<x2<<x2020,0=y2020<y2019<<y1. \begin{aligned} 0 &= x_1 < x_2 < \dots < x_{2020}, \\ 0 &= y_{2020} < y_{2019} < \dots < y_1. \end{aligned}
Denote O=(0,0)O = (0,0), the origin. Sequentially construct edges OA1,A1A2,,A2019A2020,A2020OOA_1, A_1A_2, \dots, A_{2019}A_{2020}, A_{2020}O to form a 2021-gon CC. Tao wants to paint the entire region CC in black. On each turn, he chooses a point (x,y)(x, y), pays xyxy dollars, and paints the entire region {(x,y):0xx,0yy}\{(x', y') : 0 \le x' \le x, 0 \le y' \le y\} in black. Prove that Tao can always paint the entire region CC in black while spending at most 4C4|C| dollars, where C|C| is the area of CC.

Solution

The problem can be generalized as follows: in the first quadrant, there is a continuous bounded decreasing function ff, and let CC be the region enclosed by this curve and the xx- and yy-axes. Then we can paint this region using the method described in the problem, spending at most 4C4|C| dollars. We can directly construct this painting scheme:

Figure 1

In the construction above, xx is a positive real number such that f(x)f(0)/2f(x) \ge f(0)/2 (by continuity, such an xx must exist). Then we have:
- Area of colored region No. 1 ×4=4xf(x)xf(0)+2xf(x)=the amount required when choosing (x,f(0)) plus choosing (2x,f(x)).\times 4 = 4x f(x) \ge x f(0) + 2x f(x) = \text{the amount required when choosing } (x, f(0)) \text{ plus choosing } (2x, f(x)).
- Area of colored region No. 2 ×4=4xf(2x)=the amount required when choosing (4x,f(2x)).\times 4 = 4x f(2x) = \text{the amount required when choosing } (4x, f(2x)).
- Area of colored region No. 3 ×4=8xf(4x)=the amount required when choosing (8x,f(4x)).\times 4 = 8x f(4x) = \text{the amount required when choosing } (8x, f(4x)).
- And so on in this manner.

Therefore, if we choose (x,f(0)),(2x,f(x)),(4x,f(2x)),(8x,f(4x)),(x, f(0)), (2x, f(x)), (4x, f(2x)), (8x, f(4x)), \dots in sequence, we will necessarily paint the entire region below ff (that is, CC) in black, and we have
the amount required to pay4×area of colored region4×area below f=4C \text{the amount required to pay} \le 4 \times \text{area of colored region} \le 4 \times \text{area below } f = 4|C|
Thus the construction is complete.

Figure 1

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.