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1967 AMC 12 Problem 34

Attempt Problem 34 of the 1967 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 1967 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

34.

Points D,D, E,E, FF are taken respectively on sides AB,AB, BC,BC, and CACA of triangle ABCABC so that AD:DB=1:n,AD:DB=1:n, BE:CE=1:n,BE:CE=1:n, and CF:FA=1:n.CF:FA=1:n. The ratio of the area of triangle DEFDEF to that of triangle ABCABC is:

n2−n+1(n+1)2\dfrac{n^2-n+1}{(n+1)^2}

1(n+1)2\dfrac1{(n+1)^2}

2n3(n+1)2\dfrac{2n^3}{(n+1)^2}

n3(n+1)2\dfrac{n^3}{(n+1)^2}

n(n−1)n+1\dfrac{n(n-1)}{n+1}

Answer: A
Concepts:triangle areaarea decompositionratio and proportion
Difficulty rating: 1710
Small Hint:

Subtract the three corner triangles from triangle ABCABC

Big Hint:

Each corner triangle has area ratio n(n+1)2\frac{n}{(n+1)^2}

Solution:

At each vertex, the two adjacent side fractions used by the corner triangle are 1n+1\frac{1}{n+1} and nn+1.\frac{n}{n+1}. Thus each corner triangle has area n(n+1)2\frac{n}{(n+1)^2} times [ABC].[ABC]. Therefore [DEF][ABC]=1−3n(n+1)2=n2−n+1(n+1)2. \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.

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Problem 34 in Other Years

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