Post by eric on Aug 9, 2017 10:59:27 GMT -6
Not unless you make them fit.
.
The equation of a circle with radius r centered at point (h, k) is:
(x - h) ^ 2 + (y - k) ^ 2 = r ^ 2
Let us put three circles a, b, and c as close as they can be to each other, and set all radii to 1.
Circle a is centered at point (0, 0)
Circle b is centered at point (2, 0); that is, two radii away
Circle c is centered two radii away from circle a and circle b at the same time, creating an equilateral triangle.
This puts circle c's center at point (1, sqrt(3)) because equilateral triangles have 60º corners, therefore circle c's center's height is sin(60º) * 2 = sqrt(3)
The circle that envelops all three of these (circle d) will be centered at (1, 1/sqrt(3)) because the height of the center of the equilateral triangle is 1 * tan(30º) = 1/sqrt(3)
To find the radius r of the outer circle, we can look at the point (x, y) where circle d and circle c touch:
(x - 1) ^ 2 + (y - sqrt(3)) ^ 2 = 1
(x - 1) ^ 2 + (y - 1/sqrt(3)) ^ 2 = r ^ 2
Subtracting the first equation from the second, we are left with:
(y - 1/sqrt(3)) ^ 2 - (y - sqrt(3)) ^ 2 = r ^ 2 - 1
y^2 - 2y/sqrt(3) + 1/3 - y^2 + 2y*sqrt(3) - 3 = r ^ 2 - 1
2y * (sqrt(3) - 1/sqrt(3)) - 5/3 = r ^ 2
But we know the coordinate y because it is exactly one radius above the center of circle c, which has a y coordinate of 1 + sqrt(3). Thus:
2 * (1 + sqrt(3)) * (sqrt(3) - 1/sqrt(3)) - 5/3 = r ^ 2
2 * (sqrt(3) + 3 - 1/sqrt(3) - 1) - 5/3 = r ^ 2
4 - 5/3 + 2 * (sqrt(3) - 1/sqrt(3)) = r ^ 2
7/3 + 2 * (sqrt(3) - sqrt(3)/3)) = r ^ 2
7/3 + 4*sqrt(3)/3 = r ^ 2
r ≈ 2.15
or, the hoop has to be slightly larger than twice the size of the three balls for them to fit through
an NBA regulation ball has a circumference of 29.5 inches, which means a radius of 29.5 / (2*pi) ≈ 4.7 inches
an NBA regulation hoop has a diameter of 18 inches, which means a radius of 9 inches
4.7 * 2.15 ≈ 10.1 > 9
therefore three regulation basketballs cannot fit through a regulation hoop without shenanigans
a ball would need to have a radius of 9 / 2.15 ≈ 4.2 or a circumference of 26.3 inches to fit three through easily
.
The equation of a circle with radius r centered at point (h, k) is:
(x - h) ^ 2 + (y - k) ^ 2 = r ^ 2
Let us put three circles a, b, and c as close as they can be to each other, and set all radii to 1.
Circle a is centered at point (0, 0)
Circle b is centered at point (2, 0); that is, two radii away
Circle c is centered two radii away from circle a and circle b at the same time, creating an equilateral triangle.
This puts circle c's center at point (1, sqrt(3)) because equilateral triangles have 60º corners, therefore circle c's center's height is sin(60º) * 2 = sqrt(3)
The circle that envelops all three of these (circle d) will be centered at (1, 1/sqrt(3)) because the height of the center of the equilateral triangle is 1 * tan(30º) = 1/sqrt(3)
To find the radius r of the outer circle, we can look at the point (x, y) where circle d and circle c touch:
(x - 1) ^ 2 + (y - sqrt(3)) ^ 2 = 1
(x - 1) ^ 2 + (y - 1/sqrt(3)) ^ 2 = r ^ 2
Subtracting the first equation from the second, we are left with:
(y - 1/sqrt(3)) ^ 2 - (y - sqrt(3)) ^ 2 = r ^ 2 - 1
y^2 - 2y/sqrt(3) + 1/3 - y^2 + 2y*sqrt(3) - 3 = r ^ 2 - 1
2y * (sqrt(3) - 1/sqrt(3)) - 5/3 = r ^ 2
But we know the coordinate y because it is exactly one radius above the center of circle c, which has a y coordinate of 1 + sqrt(3). Thus:
2 * (1 + sqrt(3)) * (sqrt(3) - 1/sqrt(3)) - 5/3 = r ^ 2
2 * (sqrt(3) + 3 - 1/sqrt(3) - 1) - 5/3 = r ^ 2
4 - 5/3 + 2 * (sqrt(3) - 1/sqrt(3)) = r ^ 2
7/3 + 2 * (sqrt(3) - sqrt(3)/3)) = r ^ 2
7/3 + 4*sqrt(3)/3 = r ^ 2
r ≈ 2.15
or, the hoop has to be slightly larger than twice the size of the three balls for them to fit through
an NBA regulation ball has a circumference of 29.5 inches, which means a radius of 29.5 / (2*pi) ≈ 4.7 inches
an NBA regulation hoop has a diameter of 18 inches, which means a radius of 9 inches
4.7 * 2.15 ≈ 10.1 > 9
therefore three regulation basketballs cannot fit through a regulation hoop without shenanigans
a ball would need to have a radius of 9 / 2.15 ≈ 4.2 or a circumference of 26.3 inches to fit three through easily
