1961 AMC 12 Problem 21

Question

Medians

AD

and

CE

of triangle

ABC

intersect in

M.

The midpoint of

AE

is

N.

Let the area of triangle

MNE

be

k

times the area of triangle

ABC.

Then

k

equals:

\dfrac16

\dfrac18

\dfrac19

\dfrac1{12}

\dfrac1{16}

Use coordinates or vectors with

A

as the origin

Points

E,

N,

and

M

divide the same two median directions by simple fractions

Set

A=\mathbf0,

B=\mathbf b,

and

C=\mathbf c.

Then

\begin{aligned} E&=\frac{\mathbf b}{2},\\ N&=\frac{\mathbf b}{4},\\ M&=\frac{\mathbf b+\mathbf c}{3}. \end{aligned}

A determinant calculation gives

[MNE]=\frac1{24} \left|\det(\mathbf b,\mathbf c)\right|,

while

[ABC]=\frac12 \left|\det(\mathbf b,\mathbf c)\right|.

Their ratio is

\frac{1}{12}.

Thus, the correct answer is D.

Problem 21 in other years