1950 AMC 12 第 47 题

Question

一个内接于三角形的矩形，其底边与三角形长为

b

的底边重合。若三角形的高为

h

，矩形的高

x

是矩形底边的一半，则：

A rectangle inscribed in a triangle has its base coinciding with the base

b

of the triangle. If the altitude of the triangle is

h,

and the altitude

x

of the rectangle is half the base of the rectangle, then:

x=\dfrac12h

x=\dfrac{bh}{b+h}

x=\dfrac{bh}{2h+b}

x=\sqrt{\dfrac{hb}{2}}

x=\dfrac12b

三角形内距底边高度为

x

的横截线长度为

b(1-\frac{x}{h})

The segment across the triangle at height

x

has length

b(1-\frac{x}{h})

矩形的底边长为

2x

The rectangle’s base is

2x

由相似三角形可知，在距底边高度

x

处，三角形的宽度为

b(1-\tfrac{x}{h})

，这就是内接矩形的底边。由于矩形的高是底边的一半，其底边长为

2x

。因此，

2x=b\left(1-\frac{x}{h}\right)\text{。}

两边乘以

h

并求解，得到

x(2h+b)=bh

，所以

x=\dfrac{bh}{2h+b}

。

因此，正确答案是 C。

By similarity, the width of the triangle at height

x

above its base is

b(1-\tfrac{x}{h}).

This is the base of the inscribed rectangle. Since the rectangle’s altitude is half its base, its base is

2x.

Hence

2x=b\left(1-\frac{x}{h}\right).

Multiplying by

h

and solving gives

x(2h+b)=bh,

so

x=\dfrac{bh}{2h+b}.

Thus, the correct answer is C.

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