1961 AMC 12 Problem 16

Attempt Problem 16 of the 1961 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1961 AMC 12 solutions, or check the answer key.

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16.

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)}

Let

t

be the amount removed, so the new base is

b-t

Set

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

equal to half the original area

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)},

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

Therefore, the correct answer is E.

Problem 16 in other years