Question

设 $a, b, x$ 和 $y$ 为实数，满足 $a \gt 4$ 且 $b \gt 1$ ，并且求 $a + b$ 的最小可能值。 $\begin{aligned} \frac{x^2}{a^2} + \frac{y^2}{a^2 - 16} &= \frac{(x - 20)^2}{b^2 - 1} \\ &\quad {}+ \frac{(y - 11)^2}{b^2} \\ &= 1. \end{aligned}$

Let $a, b, x,$ and $y$ be real numbers with $a \gt 4$ and $b \gt 1$ such that $\begin{aligned} \frac{x^2}{a^2} + \frac{y^2}{a^2 - 16} &= \frac{(x - 20)^2}{b^2 - 1} \\ &\quad {}+ \frac{(y - 11)^2}{b^2} \\ &= 1. \end{aligned}$ Find the least possible value of $a + b.$

答案

答案：23

知识点：椭圆三角不等式最优化

难度评级：3160

解答：

第一个椭圆有 $c^2 = a^2 - (a^2 - 16) = 16$ ，所以焦点为 $F_1 = (-4, 0)$ 和 $F_2 = (4, 0)$ 距离和为 $2a$ 。第二个椭圆以 $(20, 11)$ 为中心，长轴竖直，且 $c^2 = b^2 - (b^2 - 1) = 1$ ，所以焦点为 $G_1 = (20, 10)$ 和 $G_2 = (20, 12)$ 距离和为 $2b$ 。若 $P = (x, y)$ 在两个椭圆上，则所以 $a + b \ge 23$ 。 $\begin{aligned} 2a + 2b \\ &= (PF_1 + PG_1) \\ &\quad {}+ (PF_2 + PG_2) \\ &\ge F_1G_1 + F_2G_2 \\ &= \sqrt{24^2 + 10^2} \\ &\quad {}+ \sqrt{16^2 + 12^2} \\ &= 26 + 20 \\ &= 46, \end{aligned}$

等号要求 $P$ 同时位于线段 $\overline{F_1G_1}$ 和 $\overline{F_2G_2}$ 上。这两条线段确实相交于 $P = \left(14, \frac{15}{2}\right)$ ：此时 $PF_1 = \frac{39}{2}$ 、 $PF_2 = \frac{25}{2}$ ，所以 $2a = 32$ ， $a = 16 \gt 4$ ；且 $PG_1 = \frac{13}{2}$ 、 $PG_2 = \frac{15}{2}$ ，所以 $2b = 14$ ， $b = 7 \gt 1$ 。

因此 $a + b$ 的最小可能值为 $16 + 7 = 23$ 。

The first ellipse has $c^2 = a^2 - (a^2 - 16) = 16,$ hence foci $F_1 = (-4, 0)$ and $F_2 = (4, 0),$ with distance sum $2a.$ The second is centered at $(20, 11)$ with vertical major axis and $c^2 = b^2 - (b^2 - 1) = 1,$ hence foci $G_1 = (20, 10)$ and $G_2 = (20, 12),$ with distance sum $2b.$ If $P = (x, y)$ lies on both, then $\begin{aligned} 2a + 2b \\ &= (PF_1 + PG_1) \\ &\quad {}+ (PF_2 + PG_2) \\ &\ge F_1G_1 + F_2G_2 \\ &= \sqrt{24^2 + 10^2} \\ &\quad {}+ \sqrt{16^2 + 12^2} \\ &= 26 + 20 \\ &= 46, \end{aligned}$ so $a + b \ge 23.$

Equality requires $P$ to lie on both segments $\overline{F_1G_1}$ and $\overline{F_2G_2}.$ These segments do intersect, at $P = \left(14, \frac{15}{2}\right):$ then $PF_1 = \frac{39}{2},$ $PF_2 = \frac{25}{2},$ so $2a = 32,$ $a = 16 \gt 4;$ and $PG_1 = \frac{13}{2},$ $PG_2 = \frac{15}{2},$ so $2b = 14,$ $b = 7 \gt 1.$

Hence the least possible value of $a + b$ is $16 + 7 = 23.$

2022 AIME II 第 12 题

答案：23

难度评级：3160

解答：

其他年份的第 12 题