IMO 2023 Problem 5

> Let n be a positive integer. A Japanese triangle is made up of 1 + 2 + · · · + n circles arranged in an equilateral triangular formation, where the i-th row contains exactly i circles, each with one of them colored red. A ninja path in this triangle starts from the top row, then progresses downwards to one of the two circles directly below, ultimately finishing in the bottom row. Here is a visual of a Japanese triangle with n = 6 and an example of a ninja path featuring two red circles.

> In terms of n, determine the largest k such that every Japanese triangle has a ninja path containing at least k red circles.

The initial components are fairly straightforward: - Japanese triangle - One red circle per row - Ninja path

The final statement, however, requires some contemplation.

> "Identify the greatest k such that in every Japanese triangle, there exists a ninja path with at least k red circles."

For instance, with n = 6, the provided Japanese triangle indicates that I can identify a ninja path with k = 4 red circles. The question then arises: does this hold true for every Japanese triangle with 6 rows? If not, can we find a ninja path with k = 3 red circles?

The goal is to find the largest k that guarantees a path with k red circles exists in any configuration of the triangle. Once we establish this, we can generalize for all n, meaning that for a Japanese triangle with n = 100 rows, I should be able to assert that a ninja path must contain k = ??? red circles.

Initial Thoughts and Strategy

To discover the highest number of red circles in a ninja path for any triangle, it’s logical to examine the worst-case scenario—where locating a ninja path through numerous red circles is challenging.

The best-case situation would look like all red circles lying neatly along a single ninja path.

But what might the worst-case scenario entail?

Initially, I envisioned distributing half the red circles on one diagonal and the other half on the opposite side, like so:

A ninja path that starts left would miss all red circles on the right diagonal and vice versa. Thus, in this scenario, the maximum value of k could be approximated as n/2 + 1.

I invested considerable time attempting to prove this configuration represented the worst-case Japanese triangle. Although my attempt to validate a flawed assumption ultimately failed, it did lead me to a more effective strategy worth discussing.

Mathematics often resembles navigating a dark room, where one must feel around, bump into obstacles, and adjust direction until arriving at clarity.

At any given moment, the choice to move left or right could be guided by observing which direction has more red circles in the rows below. I tried to apply this approach to achieve the purported k value of n/2 + 1.

However, after examining various setups, I realized this strategy could fail in certain situations, as demonstrated below.

From the top, it's clear that there are more circles to the left, yet choosing the left path would yield only 3 red circles, while going to the right could lead to 6.

Consequently, I had to abandon that tactic and explore a new approach.

Starting from Scratch

In times of uncertainty, a beneficial strategy for discrete math problems is to commence from the simplest possible case (a triangle with one row). From there, we can build up progressively, aiming to identify a pattern that can be generalized and proven through mathematical induction.

• For n = 1 row, all triangles contain a ninja path with k = 1 red circle.
• For n = 2 rows, all triangles contain a ninja path with k = 2 red circles.
• For n = 3 rows, all triangles contain a ninja path with k = 2 red circles.
• For n = 4 rows, all triangles contain a ninja path with k = 3 red circles.
• For n = 5, I can minimize the number of red circles to k = 3 by placing the red circle out of reach of the one directly above it. I don’t have to adhere to my previous strategy of alternating between the ends of each row, though I could choose to do so.
• Now, n = 6 presents an interesting case. By placing red circles on alternating ends of each row, I can achieve a path with k = 4 red circles. However, there exists a configuration that only allows a path with k = 3 red circles.

I’ve finally gained insight! Placing red circles at the ends of each row does not represent the worst-case scenario. In fact, the worst case is significantly more severe, indicating that the highest k value is much less than (n/2 + 1).

The Worst Case Japanese Triangle

The worst-case scenario occurs when the red circle in the 5th row is just out of reach from the one in the 4th row, and this pattern continues. This ensures that any ninja path can only touch one red circle from rows 4, 5, 6, or 7.

Similarly, in this worst-case configuration, a ninja path can only include one red circle from rows 8 to 15, one from rows 16 to 31, and generally, one from rows 2^m to (2^(m+1) - 1) for any integer m.

Thus, the solution to the problem—the maximum number of red circles k that must exist along a ninja path—depends entirely on the largest power of 2 less than n, where n refers to the total number of rows in the Japanese triangle.

If 2^m is the largest power of 2 less than n, then there exists a ninja path with k = (m + 1) red circles. The +1 accounts for the red circle in the top row (the 2^m row).

To express this functionally in terms of n, we can utilize a base 2 logarithm:

k = ?log?(n)? + 1, where ?log?(n)? denotes the greatest integer less than or equal to log?(n).

This outcome is indeed surprising! And it is significantly more unfavorable than the (n/2 + 1) that I initially believed was the worst-case scenario.

Proof that it can’t get any worse

With the arrangement illustrated above, it's evident that there exists a Japanese triangle that lacks any ninja paths with more than k = ?log?(n)? + 1 red circles.

However, it's essential to clarify that this truly represents the worst-case scenario, and all Japanese triangles indeed possess a ninja path with k = ?log?(n)? + 1 red circles.

For this, I will employ a proof by induction on m, where 2^m is the largest power of 2 less than the number of rows n.

The statement I intend to prove is as follows:

The ‘worst-case’ Japanese triangle with (2^m - 1) rows—the triangle that minimizes the maximum number of red circles k in any ninja path—must have ninja paths with m red circles terminating at every circle in the (2^m - 1)’th row.

Here is a diagram to illustrate this concept:

To demonstrate this statement using mathematical induction, we begin with a base case of m = 1, confirming that to minimize the highest number of red circles, every circle in the 2¹ - 1 = 1st row must have a ninja path containing 1 red circle.

Next, we assume that for m = p, the triangle minimizing the maximum number of red circles k in any ninja path must have ninja paths with p red circles ending at every circle in the (2^p - 1)’th row.

We now aim to prove that for m = p + 1, the triangle minimizing the highest k in any ninja path must have paths with (p + 1) red circles ending at every circle in the (2^(p+1) - 1)’th row.

Consider the red circle in the (2^p)’th row. There exists a ninja path to this circle containing (p + 1) red circles. Below this circle is an equilateral triangle that cannot accommodate any additional red circles, as these would be reachable from above, which we want to avoid for a ninja path with more than (p + 1) red circles.

We must place one red circle in each of the remaining (2^p - 1) rows across the remaining (2^p - 1) diagonals (since placing two red circles in one diagonal would yield a ninja path with more than (p + 1) red circles).

This leads to the (2^(p+1) - 1)’th row, with every circle ending a ninja path with (p + 1) red circles.

The proof by induction is now complete.

The ‘worst-case’ Japanese triangle with (2^m - 1) rows must contain ninja paths with m red circles terminating at every circle in the (2^m - 1)’th row.

Now, by placing a red circle in the (2^m)’th row, we guarantee a ninja path with (m + 1) red circles.

The journey of this ninja has reached a pause, at least for the moment.