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?
Call the square ABCD, with AC and BD being the diagonals.
Let the 45-degree line be EF with E and AB and F on AD. EF will be parallel to BD.
Draw diagonal AC and let G be where the diagonal intersects EF.
Then we will have partitioned the square into three regions AEF, BCGE, and DCGF.
The three circles will be inside these three regions. For the two quadrilaterals note that BE and DF will not be tangent to a circle but the three other sides of each quadrilateral will be.
Now extend BC and EF to intersect at H. Similarly, extend CD and EF to intersect at I.
Then BCGE is part of triangle HCG and DCGF is part of ICG.
Now all three circles will be the incircles of AEF, HCG, and ICG. Note that the incircles of HCG and ICG are tangent to each other at a point on common side CG.
AEF, HCG, and ICG are all 45-45-90 triangles.
All three incircles will be congruent when all three triangles are congruent.
Let the side length of the square be 1, then diagonal AC will have length sqrt(2). Let x be the length of one leg.
From either triangles HCG or ICG, GC has length x
From triangle AEF, AG is its internal altitude and will have length x/sqrt(2).
From AC=AG+GC we can construct the equation sqrt(2)=x+x/sqrt(2). Then x=2*sqrt(2)-2, or x~=0.82843.
This is the distance that each endpoint of line EF is away from common corner A (with a unit length side of the square).
Edited on September 23, 2023, 7:48 pm
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