Edited by Richard Walker, Saturday 26 July 2025 at 00:45
Malfatti asked how to find three circles in a given triangle, such that each circle would be tangent to the other two cirrcles, and to two sides of the triangle. It seems to have been thought at one time that this would also be the solution to the (different) problem of packing three (not necessarily equal) circles into a triangle so as maximise the are covered by the circles.
For an equilateral triangle Malfatti's original problem is solved by the arrangement in Figure 1.
Intuitively it seems quite plausible that might at the same time maximise the area covered. There doesn't seem to be any way we can make any of the circles bigger without the others becoming smaller, and we might expect the case when the circles are equal to be optimal.
But it isn't; the arrangement of Figure 2 covers a bigger area.
Rather pleasingly the small circles are one-third the size of the big one, as illustrated by te three dotted circles.
The difference is not great, but after some trigonometry and algebra it emerges that Figure 2 is the clear winner, covering 73.9% as opposed to Figure 1 with only 72.9%.
This set me thinking: what if we replace the equilateral triangle with a square and asked for four circles that cover as much of the square as possible?
The arrangement analogous to Figure 2 appears in Figure 3.
After a bit of calculation, we find this covers 85.5%, and based on our earlier experiences we'd expect this does better than using four equal triangles.
And indeed it does, and we need no calculations; here is a "look and see" proof.
This just amounts to four half-scale copies of Figure 3, but minus the three small circles. So in each smaller square we are covering less than the 85.5% of Figure 3, and the overall percentage covered must also fall short of this value.
This is not a proof that Figure 3 is the best possible of course; only that it is better than Figure 4. Maybe some other arrangement might achieve more than 85.5%.
Malfatti Squared
Malfatti asked how to find three circles in a given triangle, such that each circle would be tangent to the other two cirrcles, and to two sides of the triangle. It seems to have been thought at one time that this would also be the solution to the (different) problem of packing three (not necessarily equal) circles into a triangle so as maximise the are covered by the circles.
For an equilateral triangle Malfatti's original problem is solved by the arrangement in Figure 1.
Intuitively it seems quite plausible that might at the same time maximise the area covered. There doesn't seem to be any way we can make any of the circles bigger without the others becoming smaller, and we might expect the case when the circles are equal to be optimal.
But it isn't; the arrangement of Figure 2 covers a bigger area.
Rather pleasingly the small circles are one-third the size of the big one, as illustrated by te three dotted circles.
The difference is not great, but after some trigonometry and algebra it emerges that Figure 2 is the clear winner, covering 73.9% as opposed to Figure 1 with only 72.9%.
This set me thinking: what if we replace the equilateral triangle with a square and asked for four circles that cover as much of the square as possible?
The arrangement analogous to Figure 2 appears in Figure 3.
After a bit of calculation, we find this covers 85.5%, and based on our earlier experiences we'd expect this does better than using four equal triangles.
And indeed it does, and we need no calculations; here is a "look and see" proof.
This just amounts to four half-scale copies of Figure 3, but minus the three small circles. So in each smaller square we are covering less than the 85.5% of Figure 3, and the overall percentage covered must also fall short of this value.
This is not a proof that Figure 3 is the best possible of course; only that it is better than Figure 4. Maybe some other arrangement might achieve more than 85.5%.