2024 AMC 12B 第 13 题

Question

实数

x

，

y

，

h

和

k

满足方程组

x^2 + y^2 - 6x - 8y = h

x^2 + y^2 - 10x + 4y = k\text{。}

求

h + k

的最小可能值。

There are real numbers

x,

y,

h,

and

k

that satisfy the system of equations

x^2 + y^2 - 6x - 8y = h

x^2 + y^2 - 10x + 4y = k.

What is the minimum possible value of

h + k?

-54

-46

-34

-16

16

将两个方程相加，把

h + k

写成关于

x

和

y

的一个表达式。

Add the two equations to get

h + k

as a single expression in

x

and

y

配方得

h + k = 2(x-4)^2

+ 2(y-1)^2 - 34

，当两个平方项都为零时最小。

Complete the square:

h + k = 2(x-4)^2

+ 2(y-1)^2 - 34,

which is smallest when both squares vanish

两式相加，得

\begin{aligned} h + k &= 2x^2 + 2y^2 - 16x - 4y \\ &= 2(x - 4)^2 \\ &\quad {}+ 2(y - 1)^2 - 34\text{。} \end{aligned}

两个平方项都非负，所以最小值在

x = 4

、

y = 1

时取得，为

h + k = -34

。

所以正确答案是 C。

Adding the equations,

\begin{aligned} h + k &= 2x^2 + 2y^2 - 16x - 4y \\ &= 2(x - 4)^2 \\ &\quad {}+ 2(y - 1)^2 - 34. \end{aligned}

Both squared terms are nonnegative, so the minimum occurs at

x = 4,

y = 1,

giving

h + k = -34.

Thus, the correct answer is C.

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