AMC12 2024 B

AMC12 2024 B · Q13

AMC12 2024 B · Q13. It mainly tests Systems of equations, Quadratic equations.

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 $h+k$?

存在实数 $x,y,h$ 和 $k$ 满足方程组\[x^2 + y^2 - 6x - 8y = h\]\[x^2 + y^2 - 10x + 4y = k\]$h+k$ 的最小可能值是多少？

(A) -54 -54

(B) -46 -46

(D) -16 -16

(E) 16 \qquad 16 \qquad

Answer

Correct choice: (C)

正确答案：（C）

Solution

Adding up the first and second equation, we get: \begin{align*} h + k &= 2x^2 + 2y^2 - 16x - 4y \\ &= 2x^2 - 16x + 2y^2 - 4y \\ &= 2(x^2 - 8x) + 2(y^2 - 2y) \\ &= 2(x^2 - 8x + 16) - (2)(16) + 2(y^2 - 2y + 1) - (2)(1) \\ &= 2(x - 4)^2 + 2(y - 1)^2 - 34 \end{align*} All squared values must be greater than or equal to $0$. As we are aiming for the minimum value, we set the two squared terms to be $0$. This leads to $\min(h + k) = 0 + 0 - 34 = \boxed{\textbf{(C)} -34}$

将第一第二个方程相加，得到： \begin{align*} h + k &= 2x^2 + 2y^2 - 16x - 4y \\ &= 2x^2 - 16x + 2y^2 - 4y \\ &= 2(x^2 - 8x) + 2(y^2 - 2y) \\ &= 2(x^2 - 8x + 16) - (2)(16) + 2(y^2 - 2y + 1) - (2)(1) \\ &= 2(x - 4)^2 + 2(y - 1)^2 - 34 \end{align*} 所有平方项必须大于或等于 $0$。为了求最小值，我们将两个平方项设为 $0$。这导致 $\min(h + k) = 0 + 0 - 34 = \boxed{\textbf{(C)} -34}$

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