1967 AMC 12 Problem 34

Question

Points

D,

E,

F

are taken respectively on sides

AB,

BC,

and

CA

of triangle

ABC

so that

AD:DB=1:n,

BE:CE=1:n,

and

CF:FA=1:n.

The ratio of the area of triangle

DEF

to that of triangle

ABC

is:

\dfrac{n^2-n+1}{(n+1)^2}

\dfrac1{(n+1)^2}

\dfrac{2n^3}{(n+1)^2}

\dfrac{n^3}{(n+1)^2}

\dfrac{n(n-1)}{n+1}

Subtract the three corner triangles from triangle

ABC

Each corner triangle has area ratio

\frac{n}{(n+1)^2}

At each vertex, the two adjacent side fractions used by the corner triangle are

\frac{1}{n+1}

and

\frac{n}{n+1}.

Thus each corner triangle has area

\frac{n}{(n+1)^2}

times

[ABC].

Therefore

\begin{aligned} \frac{[DEF]}{[ABC]} &=1-\frac{3n}{(n+1)^2}\\ &=\frac{n^2-n+1}{(n+1)^2}. \end{aligned}

Therefore, the correct answer is A.

Problem 34 in other years