1950 AMC 12 第 48 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。

48.

在一个等边三角形内部任取一点,从该点分别向三边作垂线。这三条垂线段的长度之和:

A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is:

当该点为三角形重心时最小

Least when the point is the center of gravity of the triangle

大于三角形的高

Greater than the altitude of the triangle

等于三角形的高

Equal to the altitude of the triangle

等于三角形周长的一半

One-half the sum of the sides of the triangle

当该点为三角形重心时最大

Greatest when the point is the center of gravity

答案:C
知识点:维维亚尼定理等边三角形面积分割
难度评级:1920
小提示:

将内部点与三个顶点分别连接

Join the interior point to all three vertices

大提示:

以相同的边长为底,将三个小三角形的面积相加

Add the areas of the three smaller triangles using the common side length as their bases

解答:

设等边三角形的边长为 ss,高为 hh,三个垂直距离为 d1,d2,d3d_1,d_2,d_3。用内部点将大三角形分成三个小三角形,得到 12s(d1+d2+d3)=12sh \frac12s(d_1+d_2+d_3)=\frac12sh\text{。}因此 d1+d2+d3=hd_1+d_2+d_3=h,与所选点的位置无关。

因此,正确答案是 C

Let the equilateral triangle have side length ss and altitude h,h, and let the three perpendicular distances be d1,d2,d3.d_1,d_2,d_3. Splitting the triangle at the interior point gives 12s(d1+d2+d3)=12sh. \frac12s(d_1+d_2+d_3)=\frac12sh. Therefore d1+d2+d3=h,d_1+d_2+d_3=h, independent of the selected point.

Thus, the correct answer is C.

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