1961 AMC 12 第 16 题

Question

将三角形的一条高

h

增加

m

。为了使新三角形的面积为原三角形面积的一半，对应底边

b

必须减去多少？

An altitude

h

of a triangle is increased by a length

m.

How much must be taken from the corresponding base

b

so that the area of the new triangle is one-half that of the original triangle?

\dfrac{bm}{h+m}

\dfrac{bh}{2h+2m}

\dfrac{b(2m+h)}{m+h}

\dfrac{b(m+h)}{2m+h}

\dfrac{b(2m+h)}{2(h+m)}

设减去的长度为

t

，则新底边为

b-t

Let

t

be the amount removed, so the new base is

b-t

令

\frac12(b-t)(h+m)

等于原面积的一半

Set

\frac12(b-t)(h+m)

equal to half the original area

若减去

t

，新面积应满足

\frac12(b-t)(h+m)=\frac12\left(\frac12bh\right)\text{。}

因此

b-t=\frac{bh}{2(h+m)}

，所以

\begin{aligned} t&=b-\frac{bh}{2(h+m)}\\ &=\frac{b(2m+h)}{2(h+m)} \end{aligned}\text{。}

因此，正确答案是 E。

If

t

is removed, the new area condition is

\frac12(b-t)(h+m)=\frac12\left(\frac12bh\right).

Hence

b-t=\frac{bh}{2(h+m)},

so

\begin{aligned} t&=b-\frac{bh}{2(h+m)}\\ &=\frac{b(2m+h)}{2(h+m)}. \end{aligned}

Therefore, the correct answer is E.

其他年份的第 16 题