Based upon the result from the last puzzle, draw any triangle and inscribe the largest possible rectangle. Find the maximum ratio of the area of the rectangle to the area of the triangle.
Source: Original.
Alan's Solution:
Choose the base line to be any edge of the triangle that has angles <= 90deg
at each end of it.
Choose the upper vertices of the rectangle to be the midpoints of the other 2
edges.
Draw an imaginary line from the apex to meet the base perpendicularly.
You now have 2 right-angle triangles.
In each of these right-angled triangles, the rectangle is now 1/2 the base
length by 1/2 the height.
Area of triangle = bh/2. Area of rectangle = bh/4 [b=base length, h=triangle
height]
Therefore ratio of rectangle to triangle is 0.5 for all triangles.
Proof that this is the biggest rectangle that can fit in a RA triangle:
Draw a straight line in the xy plane, equation y = mx + c.
The triangle formed has height c (for x=0) and base -c/m (for y=0)
The area of any rectangle inscribed is x(mx+c): A = mx^2 + cx
Differentiate to find a maximum. dA/dx = 2mx+c
This is zero at x = -c/2m [half the base length of -c/m)
y = m(-c/2m)+c = -c/2 + c = c/2 [half the base height of c]