2020 AMC 12A Problem 6

Below is the professionally curated solution for Problem 6 of the 2020 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 12A solutions, or check the answer key.

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Concepts:symmetry

Difficulty rating: 1270

6.

In the plane figure shown below, 33 of the unit squares have been shaded. What is the least number of additional unit squares that must be shaded so that the resulting figure has two lines of symmetry?

44

55

66

77

88

Solution:

For both symmetries, the lines must be the vertical and horizontal center lines of the 55-by-44 grid. Every shaded square then forces the squares obtained by reflecting it across each line.

The top square lies off-center, so its reflection group has 44 squares, requiring 33 more. The middle square sits on the central column, so its group has 22 squares, requiring 11 more. The bottom-right square again has a group of 4,4, requiring 33 more.

The least number of additional squares is 3+1+3=7.3 + 1 + 3 = 7.

Thus, D is the correct answer.

Problem 6 in Other Years