距离平方的最小值，“中点+三角不等式” 的破题妙招

$24$

【题目】已知$x_1^2 + y_1^2 = x_2^2 + y_2^2 = 8$，且$x_1x_2 + y_1y_2 = 4$，则$(x_1 + x_2 – 4\sqrt{2})^2 + (y_1 + y_2 – 8)^2$的最小值为_______.
【答案】$24$
【解析】如图 $A(x_1,y_1)$,$B(x_2,y_2)$,$C$ 是 $AB$ 中点，$D(2\sqrt{2},4)$,
$|AB| = \sqrt{(x_1 – x_2)^2 + (y_1 – y_2)^2}$ $= \sqrt{x_1^2 + y_1^2 + x_2^2 + y_2^2 – 2x_1x_2 – 2y_1y_2} = 2\sqrt{2}$,
$\triangle OAB$ 为正三角形,$|OC| = \sqrt{6}$,$|OD| = 2\sqrt{6}$，
$|CD| \geq |OD| – |OC| = \sqrt{6}$，
$4|CD|^2 = (x_1 + x_2 – 4\sqrt{2})^2 + (y_1 + y_2 – 8)^2 \geq 24$，
所以$(x_1 + x_2 – 4\sqrt{2})^2 + (y_1 + y_2 – 8)^2$的最小值为 $24$.