Project Euler Forum

If I understand correctly, we need to have $n$ circles named $c_1, \dots, c_n$, with $c_i$ and $c_{i+1}$ being harmonious for each $i$. For non-consecutive values of $i, j,$ the circles $c_i, c_j$ can intersect at non-grid points, and that's ok. But, if they do harmonize by accident, do we count those intersection points for the chain to be considered perfect?

The problem does not require a single chain to connect all circles, it asks that every two circles can be connected using a chain (a different chain for every pair).

Thanks, Urimend. I think now I have understood, but I am not so sure.
For me it´s probably the most confusing problem until now; I mean, problems may be very difficult, but usually at least I always understand what is the problem asking for.
Though the problem does not say that, the examples made me think that every pair of the circles should harmonise

I revised the question after pointed out it might contain spoilers.

If two circles intersect at a non-grid point, would they still be considered consonant as long as they can be connected by a chain of harmonized pairs?

Momotaro wrote: Sun Feb 08, 2026 3:26 pm if two circles (say C and D) intersect but only at non-grid points (so they do not harmonise)

Non-grid intersections are simply ignored, so the circles can and should be connected by a chain of harmonised pairs.

In the diagram with four circles, there are no unique harmony points: every harmony point is shared by at least three pairs of circles. Should I read "distinct harmony points" instead?

pjt33 wrote: Mon Feb 16, 2026 8:32 am Should I read "distinct harmony points" instead?

They want the count of the lattice points (x, y) in the plane which are harmony points of any pair of circles in the configuration.

In condition 5 is said that between two arbitrary chosen circles, there existiert a chain of Circles where one with the next harmonizes. But I cannot see that this chain shall contain all circles.
In the case that not all circles have to be included , then I can put circles of r=sqrt(5)
On the Grid points (3i,3i) i=0..n-1
What am I missing or misunderstanding?

mr66 wrote: Tue Mar 17, 2026 11:36 am In condition 5 is said that between two arbitrary chosen circles, there existiert a chain of Circles where one with the next harmonizes. But I cannot see that this chain shall contain all circles.
In the case that not all circles have to be included , then I can put circles of r=sqrt(5)
On the Grid points (3i,3i) i=0..n-1
What am I missing or misunderstanding?

Obviously this has too many points

urimend wrote: Sat Feb 07, 2026 5:59 pm The problem does not require a single chain to connect all circles, it asks that every two circles can be connected using a chain (a different chain for every pair).

I am a bit confused. The example provided of a sequences of circles, with circles adjacent in the sequence being harmonious, which is just a linear graph, then for any two points in the graph, there is obviously a path between the points.

How does this example contradict the property that

5. The circles are connected in the sense that a chain of circles can be formed between every pair of circles such that each circle harmonises with the next circle.

Do you mean that once you have connected two circles using a chain, all edges in the chain must be removed? This means a complete graph. But why not just ask for a complete graph?

I am very confused.

@thedoctar
A sequence of $n$ circles (each one connected to the next) have $2(n-1)$ harmony points.
For $n > 2$ it means too many harmony points.

So you're just saying it's not perfectly consonant but it is still consonant right?

@thedoctar, the graph whose vertices are circles and whose edges connect pairs of circles which harmonise must be a connected graph.

Thanks that is what I expected.