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?
I'll start by extending the bottom side of the rectangle to meet the sides of the triangle. Then that will create a new smaller triangle, in which we want to maximize the area of the rectangle.
But we need to know the size of the small triangle. To find that I'll start by constructing the other tangent to the circle parallel to the base of the main triangle. And then extending the sides of the main triangle to meet the new longer tangent. This will create a large triangle.
The small triangle, the main triangle and the large triangle are all equilateral triangles and all share a common vertex and line of altitude. The circle is now the incircle of the large triangle. But for an equilateral triangle, the incircle will partition the altitude into three congruent segments, where the center is and where the circle intersects the altitude internally.
Then the three triangles are in a 1:2:3 ratio. So the small triangle is exactly 1/2 scale to the main triangle. Then the small triangle will have side lengths of 1/2.
So now onto maximizing the rectangle in the small triangle. I'll just plop this onto a coordinate grid, with the base on the x-axis and vertex on the y-axis. Then the triangle has vertices of (-1/4,0) (1/4,0) and (0,sqrt(3)/4).
The upper corners of the rectangle will lay on the lines denoting the sides of the triangle: 4x+4y/sqrt(3)=1 and -4x+4y/sqrt(3)=1. Then let the base be length b. Then the corners of the rectangle can be parameterized as (-b/2,0), (b/2,0), (-b/2,sqrt(3)*(1-2b)/4) and (b/2,sqrt(3)*(1-2b)/4).
Then the area of the rectangle is b*sqrt(3)*(1-2b)/4 = sqrt(3)*(b-2b^2)/4. Differentiating this gives us sqrt(3)*(1-4b)/4 = 0. Then the max area occurs at b=1/4. The area is then sqrt(3)*(1/4-2/16)/4 = sqrt(3)/32.
Solution
