Minimum Shadow Area

Geometry Level 4

An ellipsoid has its first semi-axis of length 15 units along the unit vector u 1 = 1 9 ( 8 , 4 , 1 ) \mathbf{u_1} = \dfrac{1}{9} (8,-4,1) , and its second semi-axis of length 30, along the unit vector u 2 = 1 9 ( 4 , 7 , 4 ) \mathbf{u_2} = \dfrac{1}{9} (4, 7, -4) , and its third semi-axis of length 10 along the unit vector u 3 = 1 9 ( 1 , 4 , 8 ) \mathbf{u_3} = \dfrac{1}{9} (1, 4, 8 ) . It is positioned such that it touches the x y xy plane at the origin (that is, all of the ellipsoid lies above the x y xy plane, except for one point, the origin). The ellipsoid is subjected to uniform direction light (like sun rays) that produces a shadow of the ellipsoid on the x y xy plane. By varying the direction of the light rays, find the minimum possible area of that shadow.

Note: In a previous problem it was shown that there is a closed form expression for the shadow area and it is given by

A = π d 0 T Q e d 0 Q e n T d 0 A = \pi \dfrac{ \sqrt{\mathbf{d_0}^T Q_e \mathbf{d_0} }}{ \sqrt{ | Q_e | } | \mathbf{n}^T \mathbf{d_0}| }

where d 0 \mathbf{d_0} is the unit direction vector of the light rays, n \mathbf{n} is the normal to the projection plane, and where the ellipsoid equation is given by ( r r 0 ) T Q e ( r r 0 ) = 1 ( \mathbf{r} - \mathbf{r_0} )^T Q_e ( \mathbf{r} - \mathbf{r_0} ) = 1 .


The answer is 877.4793.

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