Steiner who? In 9-dimensional space.

Algebra Level pending

Reading row wise, these are ten points in nine dimensional space. You will need to compute the equation (in = 0 =0 form) of the surface of the minimal hypervolume hyperellipse with those ten points all on its surface. 0 0 1 1 0 0 1 0 1 1 1 1 1 1 0 1 1 0 1 0 0 0 0 0 1 0 0 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 0 0 0 0 1 0 1 1 1 1 1 1 1 0 1 0 1 0 0 0 1 0 1 0 0 1 0 1 1 1 0 1 1 0 \begin{array}{ccccccccc} 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 0 \\ \end{array}

The problem's title is a hint. Read the Steiner Ellipse articles.

Sum the coefficients of the surface's equation. Multiply by one thousand and round the result to an integer. If the real result to be rounded has an exact one-half, the round to the nearest even integer. This computation can be done the (in = 0 =0 form) by evaluating the left side of the equation with all variables set to one.

If you know what you are dong, then the problem can be solved without calculus.


The answer is 1625.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

The equation is x 1 2 + x 2 x 1 2 + x 3 x 1 + x 6 x 1 + 11 x 7 x 1 8 + 9 x 9 x 1 8 − 13 x 1 4 − x 4 x 1 8 − x 5 x 1 8 − 9 x 8 x 1 8 + x 2 2 8 + 5 x 3 2 8 + x 4 2 4 + x 5 2 8 + 3 x 6 2 8 + 3 x 7 2 4 + x 8 2 2 + 3 x 9 2 8 + 3 x 2 x 3 8 + x 3 x 5 8 + x 2 x 6 4 + 3 x 3 x 6 8 − 2 x 6 + 3 x 2 x 7 8 + 3 x 3 x 7 8 + x 4 x 7 8 + x 6 x 7 − x 3 x 8 + 3 x 4 x 8 8 + 7 x 8 4 + x 2 x 9 4 + x 3 x 9 2 + 5 x 6 x 9 8 + 7 x 7 x 9 8 − 2 x 9 − x 6 x 8 2 − 7 x 2 8 − 11 x 3 8 − x 2 x 4 8 − 5 x 3 x 4 8 − x 4 x 5 8 − 23 x 7 8 − 3 x 2 x 8 8 − x 5 x 8 8 − 5 x 7 x 8 8 − 5 x 8 x 9 8 + 3 \mathbf{x}_1^2+\frac{\mathbf{x}_2 \mathbf{x}_1}{2}+\mathbf{x}_3 \mathbf{x}_1+\mathbf{x}_6 \mathbf{x}_1+\frac{11 \mathbf{x}_7 \mathbf{x}_1}{8}+\frac{9 \mathbf{x}_9 \mathbf{x}_1}{8}-\frac{13 \mathbf{x}_1}{4}-\frac{\mathbf{x}_4 \mathbf{x}_1}{8}-\frac{\mathbf{x}_5 \mathbf{x}_1}{8}-\frac{9 \mathbf{x}_8 \mathbf{x}_1}{8}+\frac{\mathbf{x}_2^2}{8}+\frac{5 \mathbf{x}_3^2}{8}+\frac{\mathbf{x}_4^2}{4}+\frac{\mathbf{x}_5^2}{8}+\frac{3 \mathbf{x}_6^2}{8}+\frac{3 \mathbf{x}_7^2}{4}+\frac{\mathbf{x}_8^2}{2}+\frac{3 \mathbf{x}_9^2}{8}+\frac{3 \mathbf{x}_2 \mathbf{x}_3}{8}+\frac{\mathbf{x}_3 \mathbf{x}_5}{8}+\frac{\mathbf{x}_2 \mathbf{x}_6}{4}+\frac{3 \mathbf{x}_3 \mathbf{x}_6}{8}-2 \mathbf{x}_6+\frac{3 \mathbf{x}_2 \mathbf{x}_7}{8}+\frac{3 \mathbf{x}_3 \mathbf{x}_7}{8}+\frac{\mathbf{x}_4 \mathbf{x}_7}{8}+\mathbf{x}_6 \mathbf{x}_7-\mathbf{x}_3 \mathbf{x}_8+\frac{3 \mathbf{x}_4 \mathbf{x}_8}{8}+\frac{7 \mathbf{x}_8}{4}+\frac{\mathbf{x}_2 \mathbf{x}_9}{4}+\frac{\mathbf{x}_3 \mathbf{x}_9}{2}+\frac{5 \mathbf{x}_6 \mathbf{x}_9}{8}+\frac{7 \mathbf{x}_7 \mathbf{x}_9}{8}-2 \mathbf{x}_9-\frac{\mathbf{x}_6 \mathbf{x}_8}{2}-\frac{7 \mathbf{x}_2}{8}-\frac{11 \mathbf{x}_3}{8}-\frac{\mathbf{x}_2 \mathbf{x}_4}{8}-\frac{5 \mathbf{x}_3 \mathbf{x}_4}{8}-\frac{\mathbf{x}_4 \mathbf{x}_5}{8}-\frac{23 \mathbf{x}_7}{8}-\frac{3 \mathbf{x}_2 \mathbf{x}_8}{8}-\frac{\mathbf{x}_5 \mathbf{x}_8}{8}-\frac{5 \mathbf{x}_7 \mathbf{x}_8}{8}-\frac{5 \mathbf{x}_8 \mathbf{x}_9}{8}+3 .

The answer is 13 8 \frac{13}{8}

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