Parallel Lines
How close can parallel lines that trisect a square be?
Parallel lines cut a square into 3 pieces with equal areas. If the edge of the square is 3 units, what is the minimum possible distance between the parallel lines?
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This is an easy question, so an exact answer with a proof will probably be needed to get best answer. Note the 3 pieces do not have to be congruent, but they must have the same area.
You would think the answer would be 1 unit apart (making the lines parallel to the sides), but you can get closer if you make the lines parallel to a diagonal.
Let’s focus on half the square cut diagonally. That would make a 45-45-90 triangle.
The area would be 4.5 You want to add a parallel line a certain distance parallel to the hypotenuse such that the area of the triangle would be 2/3 of the total (3 sq. units) and the trapezoid area is 1/3 of the total (1.5 sq. units).
Call the distance from the *corner* to the parallel line ‘x’. This is also the height of the smaller 45-45-90 triangle of area 3 sq. units.
This smaller triangle will have a height of x and a base of 2x. So the area will be:
1/2 * 2x * x = 3
Solve for x:
x^2 = 3
x = sqrt(3)
That’s the distance from the corner to each parallel line.
The total diagonal would be 3 * sqrt(2) from which you need to subtract two times the prior result.
3 sqrt(2) – 2 sqrt(3)
That’s the exact value, but you can get an approximate value if you like:
Answer:
3 sqrt(2) – 2 sqrt(3)
(approx. 0.778539072 units)
Edit: You could prove this with calculus but you can use an intuitive geometric proof too. First, because of symmetry the parallel lines will have to form two congruent regions for the outside areas. But the middle area may change. Imagine a line drawn through the center of the square. Perpendicular to that, draw the two parallel lines. Align the whole “structure” so that the lines are parallel to the sides. This is actually the worst case where the lines are 1 unit apart. Now imagine rotating this structure. The lines would get closer together. The next interesting point would be when each line is touching an opposite corner. This is smaller than the prior case, but still not minimum. If you continue rotating the structure you get to the diagonal arrangment. Here you have maximized the area of the outside triangles. You can see that if you start turning it more, you get to a symmetric arrangement equivalent to one with the lines further apart. So the diagonal arrangement is the best.
Junior boys – Parallel Lines
