Solving An Insanely Hard Problem For High School Students

Summary

In this video, Presh Talwalkar introduces a challenging problem from the International Mathematical Olympiad (IMO) 2019, where participants solve six complex math problems over two days. He discusses the problem involving a functional equation and guides viewers through the solution process. Starting with simple substitutions and logical deductions, he demonstrates how to narrow down the solution to two possibilities: f(x) = 0 or f(x) = 2x + n. The video highlights the difficulty of the problem, providing insight into the high level of skill needed to compete in the IMO while breaking down a solution step-by-step.

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Q & A

What is the International Mathematical Olympiad (IMO)?

The International Mathematical Olympiad (IMO) is an annual competition for pre-college students, where more than 100 countries participate, each sending a team of six elite math competitors.

How is the IMO structured in terms of questions and time?

On each day of the IMO, students solve three questions in 4 1/2 hours. Each problem is worth 7 points, giving a total of 42 points for the entire competition.

What was the mean score at the IMO in 2019?

The individual mean score at the IMO in 2019 was about 16 points.

What was the challenge level of the problem discussed in the video for most competitors?

The problem discussed in the video was considered quite challenging for most competitors, but for around 60% of them, it was an easy problem.

What is the key equation in the problem from Day 1 of the IMO?

The key equation in the problem is: f(2a) + 2f(b) = f(f(a + b)), where f is a function mapping integers to integers.

What substitution is made when a = 0 in the problem?

When a = 0, the equation becomes: f(0) + 2f(b) = f(f(0 + b)), which simplifies to f(0) + 2f(b) = f(f(b)).

What conclusion is drawn after trying several special values for 'a'?

After testing several values of 'a', the equation leads to the realization that f(x) can be written as a linear function of the form f(x) = mx + n.

How does the video suggest solving for m and n in the linear function form?

To find m and n, the coefficients of like terms are matched in the equation derived from substituting the linear function into the original problem. This leads to two possible cases for m and n.

What are the two possible solutions for f(x)?

The two possible solutions are: f(x) = 0 (the trivial solution) and f(x) = 2x + n, where n is any integer.

How can the validity of the two solutions be verified?

The validity of both solutions can be verified by substituting them back into the original equation, where they both satisfy the equation, confirming they are correct.