On September 3rd, BoJone and nine classmates went to Yunfu to participate in this year’s Guangdong Province Mathematics Competition Preliminaries. The scenes of setting off together, joking, competing, and eating together are still vivid in my mind, leaving a lasting impression. Indeed, the feeling of fighting side by side is wonderful! The math results came out on the 9th. Unfortunately, the policy changed this year; I was informed that there were only three spots for the entire city to qualify for the finals. Consequently, I was the only one from Xinxing to enter the finals (the other two are said to be from Luoding, and the three of us tied for first place). It’s a bit disappointing; I suppose they want to filter out those who participate solely for utilitarian reasons...
The preliminary questions in Guangdong this year were unprecedentedly simple, whether compared to other regions in the country or to the previous year’s questions. However, I still didn’t perform as well as I hoped. According to my estimation, I might get at most 68 points on the 120-point paper, so BoJone’s basic skills are not very optimistic. After returning from the exam in Yunfu, I discussed the questions with my classmates and derived some very interesting results. The process was truly delightful! Below are several ingenious solutions to the penultimate preliminary question for everyone to enjoy. These solutions were completed by myself and Wu Zeqi (known as "Rabbit" or "Divine Rabbit," a name that fits his clever and cute personality).
Problem:
In a line segment, randomly select two points to cut the segment into three pieces. Find the probability that these three segments can form a triangle.
The problem is easy to understand, and the answer is also simple: 1/4. The conventional approach to this type of problem is linear programming, but linear programming is an algebraic idea belonging to the realm of analytical geometry. Since the answer 1/4 is so concise, we felt that the linear programming approach was almost "ugly." We wanted to find a clearer, more direct line of thought and hoped to move closer to a pure geometric approach. It was because of this idea that the following three solutions were born.
I. Circular and Symmetry Idea
Connect the ends of the line segment to be cut to form a circle. Let A be the connection point, and let the other two cut points be C and D. As long as A, C, and D are not within the same semicircle, the circle can be "straightened" into \Delta ACD. As shown in the figure, if AB is the diameter, then C and D must be separated by AB. The probability of this is 1/2 (the probability of C and D both being below AB is 1/4, and the same applies to the area above).
If we randomly select a point C on the circle, the valid region for D is the minor arc BB'. Simultaneously, we can find a corresponding point C' such that the invalid region for D is the minor arc AA', where AC = BC' and BB' = AA'. Thus, within the 1/2 probability, the cases where a triangle can be formed and the cases where it cannot are in one-to-one correspondence. This indicates that their probabilities are equal, meaning each accounts for half. Therefore, the probability of forming a triangle is P(\Delta) = 1/2 \cdot 1/2 = 1/4.
II. Ellipse and Circle Idea
For a triangle with a fixed perimeter, if two points are fixed and the third point moves freely, the resulting locus is an ellipse. Based on this principle, we draw concentric circles using the length of the segment and half of its length as radii. Suppose the segment AD is cut into three pieces: AB, BC, and CD. Let the segment AB coincide with the diameter MN, with its midpoint coinciding with point O. Then, only when point C is inside the small circle can we find a corresponding AB segment to form a triangle (by finding the corresponding ellipse). When point C is outside the small circle but inside the large circle, it is impossible to form a triangle of a fixed perimeter with segment AB. Thus, the probability of forming a triangle is equal to S_{\text{small circle}} / S_{\text{large circle}} = 1/4.
III. Ternary Coordinate System Idea
The idea of a ternary coordinate system is very representative, though its origin is not mathematics but geography. Indeed, the first time I saw it was on a geography exam, in a chart describing the proportions of young, middle-aged, and elderly populations in a country. These three proportions have a characteristic: their sum equals 1. The triangle discussed in this article has a similar property: the sum of the three sides is a constant value. This gave me some inspiration.
As shown in the figure, construct an equilateral triangle ABC with the segment length as its side length, and mark points D, E, and F as shown. It can be seen that PD + PE + PF = AB, and PD = AF, PF = CE, PE = BD. To form a triangle, it is only necessary to satisfy the condition that the sum of any two sides is greater than the third side, which means each side must be less than 1/2 AB. From the figure, it is evident that if any of AF, CE, or BD is greater than AB/2, then point P will fall outside the triangle GHI. Therefore, to form a triangle, point P must fall inside triangle GHI. Thus, the probability of forming a triangle is: P(\Delta) = \frac{S_{\Delta GHI}}{S_{\Delta ABC}} = 1/4
The solutions are now fully introduced.
The above are the three solutions derived by our classmates during the process of exploring the problem. Each solution attempts to stay as close to pure geometry as possible and make the line of thought as clear as possible. While appreciating the concise beauty of mathematics, we hope to demonstrate the necessity of these ideas. Of course, this is just the tip of the iceberg, and we hope these ideas serve as a "brick to attract jade." If any readers have better solutions, you are welcome to exchange them. Here, Scientific Space wishes everyone a Happy Mid-Autumn Festival!
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