1994 AMC 8 Problem 12

Below is the professionally curated solution for Problem 12 of the 1994 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1994 AMC 8 solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:area decompositionarea ratio

Difficulty rating: 820

12.

Each of the three large squares shown is the same size. Segments that intersect the sides of the squares intersect at the midpoints of the sides. How do the shaded areas of these squares compare?

The shaded areas in all three are equal.

Only the shaded areas of II and IIII are equal.

Only the shaded areas of II and IIIIII are equal.

Only the shaded areas of IIII and IIIIII are equal.

The shaded areas of I,II,I, II, and IIIIII are all different.

Solution:

In square II,II, 11 of the 44 equal cells is shaded, so 14\tfrac14 of it is shaded.

Square II breaks into 88 equal triangles with 22 shaded, and square IIIIII breaks into 1616 equal triangles with 44 shaded; each of these equals 28=416=14.\tfrac{2}{8} = \tfrac{4}{16} = \tfrac14.

Since every figure has exactly 14\tfrac14 shaded, the shaded areas are all equal.

Thus, the correct answer is A .

Problem 12 in Other Years

1985 AMC 8 · 1986 AMC 8 · 1987 AMC 8 · 1988 AMC 8 · 1989 AMC 8 · 1990 AMC 8 · 1991 AMC 8 · 1992 AMC 8 · 1993 AMC 8 · 1995 AMC 8 · 1996 AMC 8 · 1997 AMC 8 · 1998 AMC 8 · 1999 AMC 8 · 2000 AMC 8 · 2001 AMC 8 · 2002 AMC 8 · 2003 AMC 8 · 2004 AMC 8 · 2005 AMC 8 · 2006 AMC 8 · 2007 AMC 8 · 2008 AMC 8 · 2009 AMC 8 · 2010 AMC 8 · 2011 AMC 8 · 2012 AMC 8 · 2013 AMC 8 · 2014 AMC 8 · 2015 AMC 8 · 2016 AMC 8 · 2017 AMC 8 · 2018 AMC 8 · 2019 AMC 8 · 2020 AMC 8 · 2022 AMC 8 · 2023 AMC 8 · 2024 AMC 8 · 2025 AMC 8 · 2026 AMC 8