Consider all rhombuses with side length 2 whose centers are the origin and vertices lie on the coordinate axes. Find the area that envelopes these rhombuses.
Later correction: (after reading armando's correct solution):
Everything below is for a rhombus of side 1. So, the correct answer for a side=2 rhombus is 4 times greater: A = 3 pi/2
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Solution: A=3 pi/8, without a proof (via research).
I think the rhombus' envelope is an astroid:
x^(2/3) + y^(2/3) = c^(2/3), with, here, c=1
This problem is akin to the sliding ladder problem, but with the question here being not the shape traveled by the midpoint of the ladder (easily shown to be a circle), but rather the outer envelope of the shape it fills while sliding.
To see this, see "5th Problem, Generalization" here.
The integration of the area within an astroid as expressed in parametric form:
x=cos^3 t, y=sin^3 t
is done here etc. etc. (somewhat tediously), and the answer comes out:
Area = 3 pi/8
The part I worked on, by didn’t get, was motivating the astroid. The method I thought of was to place a right triangle of hypotenuse=1 attached to the x and y-axes in the first quadrant, making an arbitrary angle phi with the x-axis. Then run a line to (r, theta) from the origin to intersect the hypotenuse.
Express r as a function of theta and phi and then maximize r using phi. (i.e., set d_r/d_phi = 0)
This gives phi_max and r_max and thus r(theta). This should come out as :
r_max(theta) = sec (theta) / [1+tan(theta)^(3/2)]^(3/2) ]
as per Wolfram "MathWorld" here
This is the third form for an astroid, as Wolfram shows. Note - Wolfram starts from the parametric form, while I was trying from the problem posed.
Edited on March 13, 2019, 12:07 am
research to possible solution
