1952 AMC 12 Problem 49

Question

In the figure,

CD,

AE,

and

BF

are one-third of their respective sides. It follows that

AN_2:N_2N_1:N_1D=3:3:1,

and similarly for lines

BE

and

CF.

Then the area of triangle

N_1N_2N_3

is:

\dfrac1{10}\triangle ABC

\dfrac19\triangle ABC

\dfrac17\triangle ABC

\dfrac16\triangle ABC

None of these

Area ratios are affine-invariant, so choose convenient coordinates for triangle

ABC

Locate

D,E,F

by the one-third conditions, intersect the three cevians, and compare the two areas with determinants

Use

A=(0,1),

B=(0,0),

C=(1,0).

Then

\begin{aligned} D&=\left(\frac23,0\right),\\ E&=\left(\frac13,\frac23\right),\\ F&=\left(0,\frac13\right). \end{aligned}

Intersecting

AD,BE,CF

in pairs gives

\begin{aligned} N_1&=\left(\frac47,\frac17\right),\\ N_2&=\left(\frac27,\frac47\right),\\ N_3&=\left(\frac17,\frac27\right). \end{aligned}

The determinant area formula gives

\begin{aligned} [N_1N_2N_3]&=\frac1{14},\\ [ABC]&=\frac12. \end{aligned}

Hence

\frac{[N_1N_2N_3]}{[ABC]}=\frac{1}{7}.

Thus, the correct answer is C.

Problem 49 in other years