The base of a unit equilateral triangle is oriented horizontally. The center of a semicircle is at the midpoint of this base, and the semicircle is tangent to the other two sides of the triangle. Atop the semicircle sits a rectangle whose horizontally oriented base is tangent to the semicircle and whose upper two corners are on the other two sides of the triangle.
What is the maximum area of the rectangle?
Bonus: For extra fun, first guess whether the maximal rectangle will be flatter than a square or taller than a square. Did you guess correctly?
Upon reading the puzzle for the first time I predict it's flatter than a square.
Let's work on the left hand side of the diagram:
Place the lower-left corner of the triangle at the origin.
The left side of the triangle has the equation
y = sqrt(3) * x
The semicircle's center is at (.5, 0) and its topmost point is at (.5, sqrt(3)/4). The apex of the triangle is at (.5, sqrt(3)/2).
Let the rectangle's height be h, and its width, w.
The left vertical side of the rectangle is at x = .5-w/2 and its top is at the same x value. The rectangle's height, h = sqrt(3)/4-sqrt(3)*w/2.
Then w*h = w * (sqrt(3)/4-sqrt(3)*w/2), whose derivative is
sqrt(3)/4 - sqrt(3)*w
Setting sqrt(3)*w = sqrt(3)/4 gives
w=1/4
and h = sqrt(3)/4-sqrt(3)/8 = sqrt(3)/8
so the area is sqrt(3)/32. (approx. 0.0541265877365274)
And w = 1/4 > h=sqrt(3)/8=~.2165 in agreement with my initial guess.
|
|
Posted by Charlie
on 2023-08-08 14:15:09 |
solution
