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2024 AMC 12B 第 20 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

20.

设 AA，BB 和 CC 是平面上的点，且 AB=40AB = 40、AC=42AC = 42。令 xx 为从 AA 到 BC‾\overline{BC} 中点的线段长度。定义函数 ff，令 f(x)f(x) 为 △ABC\triangle ABC 的面积。那么 ff 的定义域是开区间 (p,q)(p, q)，且 f(x)f(x) 的最大值 rr 在 x=sx = s 时取得。求 p+q+r+sp + q + r + s。

Suppose A,A, B,B, and CC are points in the plane with AB=40AB = 40 and AC=42,AC = 42, and let xx be the length of the line segment from AA to the midpoint of BC‾.\overline{BC}. Define a function ff by letting f(x)f(x) be the area of △ABC.\triangle ABC. Then the domain of ff is an open interval (p,q),(p, q), and the maximum value rr of f(x)f(x) occurs at x=s.x = s. What is p+q+r+s?p + q + r + s?

909909

910910

911911

912912

913913

答案：C
知识点：中线（几何）三角不等式最优化
难度评级：2110
小提示：

设 a=BCa = BC，则中线满足 x2=2⋅402+2⋅422−a24x^2 = \dfrac{2\cdot 40^2 + 2\cdot 42^2 - a^2}{4}；三角形不等式 2<a<822 \lt a \lt 82 决定定义域。

The median satisfies x2=2⋅402+2⋅422−a24,x^2 = \dfrac{2\cdot 40^2 + 2\cdot 42^2 - a^2}{4}, where a=BC;a = BC; the triangle inequality 2<a<822 \lt a \lt 82 determines the domain

大提示：

面积在 ∠A=90∘\angle A = 90^\circ 时最大；求该最大面积和对应的 xx。

The area is largest when ∠A=90∘;\angle A = 90^\circ; find that maximum area and the corresponding xx

解答：

设 a=BCa = BC。中线长度满足 x2x^2 =2⋅1600+2⋅1764−a24= \dfrac{2\cdot 1600 + 2\cdot 1764 - a^2}{4} =6728−a24= \dfrac{6728 - a^2}{4}。三角形不等式要求 2<a<822 \lt a \lt 82，即 4<a2<67244 \lt a^2 \lt 6724，从而 1<x<411 \lt x \lt 41。所以 (p,q)=(1,41)(p, q) = (1, 41)。

在 AB=40AB = 40、AC=42AC = 42 固定时，面积为 12⋅40⋅42sin⁡A\tfrac12\cdot 40\cdot 42\sin A，当 ∠A=90∘\angle A = 90^\circ 时最大，得 r=840r = 840。此时 a2=402+422=3364a^2 = 40^2 + 42^2 = 3364，x2=6728−33644=841x^2 = \dfrac{6728 - 3364}{4} = 841，所以 s=29s = 29。

因此 p+q+r+sp + q + r + s =1+41+840+29= 1 + 41 + 840 + 29 =911= 911。

所以正确答案是 C。

Let a=BC.a = BC. The median length gives x2x^2 =2⋅1600+2⋅1764−a24= \dfrac{2\cdot 1600 + 2\cdot 1764 - a^2}{4} =6728−a24.= \dfrac{6728 - a^2}{4}. The triangle inequality requires 2<a<82,2 \lt a \lt 82, i.e. 4<a2<6724,4 \lt a^2 \lt 6724, which translates to 1<x<41.1 \lt x \lt 41. So (p,q)=(1,41).(p, q) = (1, 41).

With AB=40AB = 40 and AC=42AC = 42 fixed, the area 12⋅40⋅42sin⁡A\tfrac12\cdot 40\cdot 42\sin A is largest when ∠A=90∘,\angle A = 90^\circ, giving r=840.r = 840. Then a2=402+422=3364,a^2 = 40^2 + 42^2 = 3364, so x2=6728−33644=841,x^2 = \dfrac{6728 - 3364}{4} = 841, i.e. s=29.s = 29.

Thus p+q+r+sp + q + r + s =1+41+840+29= 1 + 41 + 840 + 29 =911.= 911.

Thus, the correct answer is C.

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