1996 AMC 12 Problem 19

Attempt Problem 19 of the 1996 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 1996 AMC 12 solutions, or check the answer key.

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

19.

The midpoints of the sides of a regular hexagon ABCDEFABCDEF are joined to form a smaller hexagon. What fraction of the area of ABCDEFABCDEF is enclosed by the smaller hexagon?
12\frac12
33\frac{\sqrt3}{3}
23\frac23
34\frac34
32\frac{\sqrt3}{2}
Answer: D
Concepts:similar polygonsarea ratios
Small hint:
Compare the distance between adjacent side midpoints with the original side length
Big hint:
The smaller and larger hexagons are similar, so square their side-length ratio
Solution:
Two adjacent midpoints and their shared vertex form a triangle with two sides s2\frac{s}{2} and included angle 120∘.120^\circ. Its opposite side has squared length s24+s24−2(s2)2cos⁡120∘=3s24. \begin{aligned} \frac{s^2}{4}+\frac{s^2}{4} &-2\left(\frac{s}{2}\right)^2\cos120^\circ\\ &=\frac{3s^2}{4}. \end{aligned} Thus the smaller hexagon has scale factor 32,\frac{\sqrt3}{2}, and its area ratio is 34.\frac{3}{4}. The correct answer is D.
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