Minimum Distance Between Two Ellipsoids

Calculus Level 5

In the x y z xyz coordinate system, Ellipsoid 1 has its center at point ( 0 , 0 , 0 ) (0,0,0) , and semi-axes coinciding with the vectors ( 1 , 1 , − 2 ) (1,1,-2) , ( − 1 , 1 , 0 ) (-1,1,0) , and ( 1 , 1 , 1 ) (1,1,1) . Ellipsoid 2 has its center at point ( 3 , 2 , 5 ) (3,2,5) , and semi-axes coinciding with the vectors ( 3 , 0 , 1 ) (3,0,1) , ( − 1 , 1 , 3 ) (-1,1,3) , and ( − 1 , − 10 , 3 ) (-1,-10,3) .

We want to find the minimum distance d ∗ d^* , between a point P = ( x 1 , y 1 , z 1 ) P =(x_1, y_1, z_1 ) on Ellipsoid 1 and a point Q = ( x 2 , y 2 , z 2 ) Q = (x_2, y_2, z_2 ) on Ellipsoid 2. If the minimum occurs with P ∗ = ( x 1 ∗ , y 1 ∗ , z 1 ∗ ) P^* = ({x_1}^*,{y_1}^*,{z_1}^*) , and Q ∗ = ( x 2 ∗ , y 2 ∗ , z 2 ∗ ) Q^* = ( {x_2}^*,{y_2}^*,{z_2}^*) , then report the value of ⌊ 100 ( x 1 ∗ + y 1 ∗ + z 1 ∗ + x 2 ∗ + y 2 ∗ + z 2 ∗ + d ∗ ) ⌋ \lfloor 100({x_1}^*+{y_1}^*+{z_1}^* + {x_2}^*+{y_2}^*+{z_2}^* + d^* ) \rfloor , where ⌊ ⋅ ⌋ \lfloor \cdot \rfloor is the floor function; for example, ⌊ 7.6 ⌋ = 7 \lfloor 7.6 \rfloor = 7 .

Inspiration


The answer is 796.

Steven Chase by Steven Chase , 37, USA

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1 solution

I checked that the axis sets were orthogonal. They are.

The blue line connects the objects' centers. The red line is the closest approach.

po = { 0 , 0 , 0 } p1 = { 1 , 1 , − 2 } p2 = { − 1 , 1 , 0 } p3 = { 1 , 1 , 1 } qo = { 3 , 2 , 5 } q1 = { 3 , 0 , 1 } q2 = { − 1 , 1 , 3 } q3 = { − 1 , − 10 , 3 } \text{po}=\{0,0,0\}\\ \text{p1}=\{1,1,-2\}\\ \text{p2}=\{-1,1,0\}\\ \text{p3}=\{1,1,1\}\\\\ \text{qo}=\{3,2,5\}\\ \text{q1}=\{3,0,1\}\\ \text{q2}=\{-1,1,3\}\\ \text{q3}=\{-1,-10,3\}

SquaredEuclideanDistance [ p1 cos ⁡ ( u1 ) cos ⁡ ( v1 ) + p2 cos ⁡ ( u1 ) sin ⁡ ( v1 ) + p3 sin ⁡ ( u1 ) + po , q1 cos ⁡ ( u2 ) cos ⁡ ( v2 ) + q2 cos ⁡ ( u2 ) sin ⁡ ( v2 ) + q3 sin ⁡ ( u2 ) + qo ] \text{SquaredEuclideanDistance}[\text{p1} \cos (\text{u1}) \cos (\text{v1})+\text{p2} \cos (\text{u1}) \sin (\text{v1})+\text{p3} \sin (\text{u1})+\text{po},\text{q1} \cos (\text{u2}) \cos (\text{v2})+\text{q2} \cos (\text{u2}) \sin (\text{v2})+\text{q3} \sin (\text{u2})+\text{qo}]

The expression:

( cos ⁡ ( u1 ) cos ⁡ ( v1 ) − cos ⁡ ( u1 ) sin ⁡ ( v1 ) + sin ⁡ ( u1 ) − 3 cos ⁡ ( u2 ) cos ⁡ ( v2 ) + cos ⁡ ( u2 ) sin ⁡ ( v2 ) + sin ⁡ ( u2 ) − 3 ) 2 + ( cos ⁡ ( u1 ) cos ⁡ ( v1 ) + cos ⁡ ( u1 ) sin ⁡ ( v1 ) + sin ⁡ ( u1 ) − cos ⁡ ( u2 ) sin ⁡ ( v2 ) + 10 sin ⁡ ( u2 ) − 2 ) 2 + ( − 2 cos ⁡ ( u1 ) cos ⁡ ( v1 ) + sin ⁡ ( u1 ) − cos ⁡ ( u2 ) cos ⁡ ( v2 ) − 3 cos ⁡ ( u2 ) sin ⁡ ( v2 ) − 3 sin ⁡ ( u2 ) − 5 ) 2 (\cos (\text{u1}) \cos (\text{v1})-\cos (\text{u1}) \sin (\text{v1})+\sin (\text{u1})-3 \cos (\text{u2}) \cos (\text{v2})+\cos (\text{u2}) \sin (\text{v2})+\sin (\text{u2})-3)^2\\ +(\cos (\text{u1}) \cos (\text{v1})+\cos (\text{u1}) \sin (\text{v1})+\sin (\text{u1})-\cos (\text{u2}) \sin (\text{v2})+10 \sin (\text{u2})-2)^2\\ +(-2 \cos (\text{u1}) \cos (\text{v1})+\sin (\text{u1})-\cos (\text{u2}) \cos (\text{v2})-3 \cos (\text{u2}) \sin (\text{v2})-3 \sin (\text{u2})-5)^2

Numerically minimizing that expression over the variables u 1 , v 1 , u 2 u1, v1, u2 and v 2 v2 gives the solution: d = 1.06351294965, u1 = 2.0839598080673123, v1 -= 0.4114553320267924, u2 = 0.09222339116300944\text{ and }v2 -> -2.4581210966219653). The closest point on the smaller object is { 0.61758 , 0.224887 , 1.77112 } \{0.61758,0.224887,1.77112\} . The closest point on the larger object is { 1.22044 , 0.450268 , 2.61777 } \{1.22044,0.450268,2.61777\} .

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