# High School Mathematics in China: A Challenging Reality

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## Chapter 1: Introduction to Gao Kao Math

It's widely recognized that high school mathematics in China presents significant challenges. In this article, we will explore the concluding problem from the Gao Kao math exam held in June 2024. This particular exam, known as the New Gao Kao Paper 1, was undertaken by approximately 5 million students across half of China last week.

The final problem is undoubtedly difficult, yet it offers a fascinating mathematical challenge focused on arithmetic sequences. Below, you will find the original Chinese text followed by an English translation. As we work through this problem, you’ll notice that while part 3 is indeed quite tough, the earlier sections provide valuable hints to assist students in arriving at a solution.

Feel free to revisit the problem multiple times to fully grasp it. Additionally, I have created a video discussing this problem for those who prefer a visual explanation.

### Section 1.1: Understanding Part 1

Part 1 primarily assesses our ability to comprehend the definitions presented in the question. Given a 6-term arithmetic sequence, the goal is to eliminate 2 terms, leaving a 4-term arithmetic sequence. This implies that the four remaining terms must be consecutive. For instance, by removing the first two terms, the last four terms will still form an arithmetic sequence.

Similarly, one could remove the last two terms or take out the first and last terms, resulting in four middle terms. Thus, the possible pairs of (i, j) for Part 1 are (1,2), (5,6), or (1,6).

### Section 1.2: Analyzing Part 2

In Part 2, we encounter a proof question. Before tackling it, it’s essential to recognize that we can simplify our approach by considering the sequence of natural numbers 1, 2, 3, … (4m+2). If this sequence is (i, j)-separable, then any sequence a_1, a_2, a_3, … a_(4m+2) will also be (i, j)-separable. By applying a scale factor to a sequence, we can maintain uniform differences across all terms.

Starting with m = 3, let’s examine the sequence of natural numbers from 1 to 14. Our task is to remove 2 and 13, demonstrating that the remaining terms can be divided into three arithmetic subsequences. While attempting sequences with a common difference of 3 yields success, sequences like 1, 3, 5, 7 fall short.

Having established that the sequence is (2, 13)-separable when m = 3, we must prove this holds true for all m ≥ 3. For instance, when m = 4, we have 18 terms. The four consecutive integers 15, 16, 17, and 18 already form an arithmetic sequence with a common difference of 1, thus contributing to our overall count.

## Chapter 2: Delving into Part 3

Now, we turn to Part 3, which also presents a proof question regarding the probability of a sequence being (i, j)-separable. The findings from Parts 1 and 2 will be instrumental in our analysis.

To calculate the probability, we first consider the denominator, which represents the total number of choices for (i, j) among (4m + 2) integers. This can be quantified using a binomial coefficient. The challenge lies in determining the numerator for the probability, specifically how many pairs (i, j) exist for each value of m.

By examining smaller cases, we find that for m = 1, there are 3 solution pairs. The total number of options for (i, j) between 1 and 6 is given by Bi(6, 2) = 15, leading to a probability of P1 = 3/15 = 1/5.

In Part 2, we established that once we have a solution, like (2, 13) when m = 3, it applies to all higher values of m. Thus, any elements from the solution set for m will also appear in the set for (m + 1).

To illustrate the solutions for m = 2, we start with the sequence of integers from 1 to 10. Utilizing our m = 1 solution pairs (1,2), (1,6), and (5,6), we can find additional pairs. Notably, (1,10) allows for the remaining integers to be divided into two arithmetic sequences.

This leads to a total of 6 solutions for m = 2. Furthermore, a similar approach to what we did in Part 2 reveals an extra solution of (2,9), yielding 7 solutions out of Bi(10, 2) = 45, which is greater than 1/8—an encouraging result!

To ensure this probability remains above 1/8 for any m, we adopt a systematic representation of our solutions. By plotting i on the x-axis and j on the y-axis, we visualize the solutions as points in a plane. The resulting diagrams illustrate the triangular nature of the solutions, confirming that the patterns we observe are consistent across values of m.

As we proceed, we see that solutions accumulate; for every new m, we can shift each (i, j) pair 4 places to the right, yielding additional solutions. The initial solutions, particularly those of the form (1, 4m + 2) and (2, 4m + 1), remain valid, allowing us to conclude that the number of pairs is given by (m² + m + 1).

Finally, dividing by the denominator (8m² + 6m + 1) confirms that the resulting probability exceeds 1/8, as required.

It's a commendable question, although I must say I do not envy the students who faced it under exam conditions! Congratulations to all who tackled this challenge! 💪👏

This video showcases the challenging nature of the Chinese Gao Kao exam, focusing on the most difficult math problems faced by students.

This video compares Chinese math education with American methods, highlighting the differences in teaching approaches and problem-solving techniques.