$10M AI Mathematical Olympiad Prize

I was recently in Palo Alto, and bumped into a newly founded startup (I don't remember the name unfortunately) who set themselves the grand the vision of exactly this: winning a gold medal on the international Olympiad using AI. Their plan was to build mostly on LLMs as a start, and iterate as they go. In their barebones office space, they had a poster with a countdown of the number of weeks till the event: it was 36 at the time.

It sounded interesting to wonder how far they could go with this kind of approach. I thought they were aiming for the moon: but also respected the boldness and determination. They had the funding to operate for at least a year, and were very focused to get there.

Seems like this prize will supr hundreds (or thousands) of teams competing in exactly this space. Perhaps it will have a similar effect like the $1M Netflix Prize in 2009 for recommendations algorithms!

Great, a worthy competition which being first and correctness completely matter and may the best AI model win. Hopefully DeepMind doesn’t enter otherwise they would smoke everyone else other than OpenAI.

Speaking of which, I’ll give you a 1% chance of winning with this paper from OpenAI: [0] with the MATH dataset: [1]

EDIT: Why the downvotes? I'm trying to help you here and give you a starting point to win the competition? What's wrong with helping others?

[0] https://arxiv.org/abs/2305.20050

[1] https://github.com/hendrycks/math

The IMO are a bit more involved...

e.g. IMO 2023:

> Problem 1. Determine all composite integers n > 1 that satisfy the following property: if d1, d2, . . . , dk are all the positive divisors of n with 1 = d1 Although ChatGPT 4 makes a pretty decent start on this already:

> To determine all composite integers \( n > 1 \) that satisfy the given property, we need to closely examine the condition set forth: for a composite number \( n \) with its divisors \( d_1, d_2, \ldots, d_k \) where \( 1 = d_1 Let's break down the problem:

1. *Composite Number Requirement*: We are looking for composite numbers, which are integers greater than 1 that are not prime. These numbers have more than two distinct positive divisors.

2. *Divisor Property*: The property \( d_i | (d_{i+1} + d_{i+2}) \) means that \( d_{i+1} + d_{i+2} \) is a multiple of \( d_i \).

3. *Systematic Examination*: We will analyze this property systematically for composite numbers starting from the smallest composite number (4) and going upwards.

Let's start the analysis:

- For \( n = 4 \), the divisors are \( 1, 2, 4 \). Here, \( 1 | (2 + 4) \), so 4 satisfies the condition. - For \( n = 6 \), the divisors are \( 1, 2, 3, 6 \). Here, \( 1 | (2 + 3) \) and \( 2 | (3 + 6) \), so 6 satisfies the condition. - For \( n = 8 \), the divisors are \( 1, 2, 4, 8 \). Here, \( 1 | (2 + 4) \), but \( 2 \nmid (4 + 8) \), so 8 does not satisfy the condition.

Let's implement a systematic method to identify such composite numbers up to a reasonable limit (let's say up to 100 for a start) and check if they satisfy the given condition.

The composite integers greater than 1 and up to 100 that satisfy the given property are: 4, 8, 9, 16, 25, 27, 32, 49, 64, and 81.

Each of these numbers has the characteristic that for all their positive divisors \( d_1, d_2, \ldots, d_k \) (where \( 1 = d_1 < d_2 < \cdots < d_k = n \)), the condition \( d_i \) divides \( d_{i+1} + d_{i+2} \) is true for every \( 1 \leq i \leq k - 2 \).

Tangential is there a list of Unicode characters allowed on hacker news?

﷽!

That's 300 words for saying "in order for n to satisfy the requirements, it must satisfy the requirements."

I tried an easier problem, extending Rolle's theorem to the multidimensional case with Poe.com:

Prompt: Let U be an open subset of R^n with compact closure K. Suppose f: K → R is continuous on K, differentiable on U, and satisfies f(x) = 0, for all x in K \ U. Show that there exists a in U with grad f (a) = 0.

Poe: To prove that there exists a point a in U such that grad f(a) = 0, we will utilize the concept of connectedness and the Mean Value Theorem.

Since U is an open subset of R^n with compact closure K, we know that K is a closed and bounded subset of R^n. Furthermore, since K is a compact set, it is also a connected set.

Now, let's define a new function g: K → R^n such that g(x) = (f(x), f(x), ..., f(x)) for all x in K. In other words, each component of g is equal to f(x) for all x in K.

Since f is continuous on K and differentiable on U (which is an open subset containing K), it follows that g is continuous on K and differentiable on U.

Consider the function h: K → R defined as h(x) = ||g(x)||^2 = f(x)^2 + f(x)^2 + ... + f(x)^2 = nf(x)^2.

Note that h is continuous on K since g is continuous on K and ||.||^2 is a continuous function.

Now, let's consider the set S = {x in K : f(x) = 0}. Since f(x) = 0 for all x in K \ U, we have K \ U ⊆ S. Therefore, the set S is closed.

...

Stochastic parrot indeed.