A line is drawn in a square at a 45 degree angle to the sides but not through the center. 3 circles of equal radius are packed in the square: one in the isosceles right triangle on one side of the line and two in the pentagon on the other side.

Where should the line be placed to maximize the size of the circles?

Let ABCD be the square, the cutting line is parallel to BD, closer to C than A.
A (0,0), B (1,0), C (1,1), D (0,1)
Let d be the length of each short partial segments from B to the cutting line and D to the cutting line.

For a 45-45-90 triangle of side length 's',
the radius of the largest inscribed circle is: s*(1-√2/2)
the area of the largest inscribed circle is:
pi*s^2*(3 - 2√2)/2
And since 's' = 1-d, the first circle has area:
pi*(1-d)^2*(3 - 2√2)/2

But before figuring out the pentagon, figure out the area for a few simple cases:
case 1: d=1, the equilateral triangle is non existent, the "pentagon" is the whole square.
there are 2 circles each of radius (1-√2/2)
Total area:  pi*(3-2√2) about 0.5390120844526466

case 2:  d=0,  the cutting diagonal goes through B and D.  We have one circle of radius (1-√2/2) and 2 cirlces of radius (1-√2/2)/2
Total area:  pi*(3-2√2)/2  +  2*pi*(3-2√2)/8 = (3/4)*pi*(3-2√2)
   about 0.4042590633394849

Now, back to the general case.
The pentagon is split in half by drawing the diagonal of the original square (through the center) perpendicular to the original line drawn.  This makes 2 quadrilaterals that have angles 45-90-135-90.
One of these is bounded by:
x-axis between 0 and 1; y=x; x=1 between y=0 and y=d; y = -x + d + 1
The inscribed circle's radius is r.  The center of this circle has a y coordinate of r to make it tangent with the x-axis.  The center's x coordinate is (d+1)/1.  It's radius is:
r = (d+1)/(2(1+√2))
Using Desmos helped visualize this.

For small d, the largest circle will contact the 2 diagonals and the full length side, not the short segment.  From Desmos, it looks like when d = √2 - 1, the inscribed circle will also be tangent to the short segment.  For d > √2 - 1, the above analysis is not valid.

But assuming a smaller d value, the 2 circles in the pentagon each have:
r = (d+1)/(2(1+√2))
Area = (pi*(d+1)^2) / (4(3+2√2))
multiply top and bottom by (3-2√2)
Area = (pi*(d+1)^2)(3-2√2) / 4

The total area is:
(pi/2)*(3-2√2) * [ (1-d)^2 + (d+1)^2 ]
2(pi/2)*(3-2√2) * (d^2 + 1)
... and setting the derivative to zero will result in a negative value for d.
Suggesting the largest d gives the maximum area.
So choose d = √2-1
2(pi/2)*(3-2√2) * ((√2-1)^2 + 1)
pi*(3-2√2) * (2+1-2√2 + 1)
pi*(3-2√2) * (4-2√2)
pi*(12 - 6√2  - 8√2 + 8)
pi*(20 - 14√2) = ~0.6314919375787372

Edited on September 22, 2023, 1:53 pm

Posted by Larry on 2023-09-22 11:39:35