SAT Question Everyone Got Wrong

[Albert Chan]

(12-03-2023 10:38 PM)ijabbott Wrote:  And for a circle rotating inside any shape with corners, one cannot merely subtract 1 from the quotient of the perimeter of the shape and the circumference of the circle, because the interface between the shapes is no longer continuous. Instead, for a rectangle you would need to subtract four circle radii from the perimeter of the rectangle, and divide that by the circumference of the circle.

You are right! There is a limit of bending of the tracks.

I assumed small circle does not simultaneously touch 2 "floors".
If it does, then we have to analyze the trip piecewise, to remove effect of skipped corners.
Note: this applied to small circle rotating outside irregular shape with valleys as well.

If small circle is rotating inside a square (or rectangle) with 3 times circle perimeter:
tan(90°/2) = (d/2) / (x/2) --> x = d

Revolutions = 4 * (3*pi*d/4 - d) / (pi*d) = 3 - 4/pi ≈ 1.727

If small circle is rotating inside an equilateral triangle with 3 times circle perimeter:
tan(60°/2) = (d/2) / (x/2) --> x = √3*d

Revolutions = 3 * (3*pi*d/3 - √3*d) / (pi*d) = 3 - 3*√3/pi ≈ 1.346

Without considering skipped corners, we would get (for both cases), revolutions = 3 - 1 = 2