2019 AMC 8 Problem 6

Attempt Problem 6 of the 2019 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 8 solutions, or check the answer key.

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

6.

There are 8181 grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point PP is in the center of the square. Given that point QQ is randomly chosen among the other 8080 points, what is the probability that the line PQPQ is a line of symmetry for the square?
15\displaystyle \dfrac{1}{5}
14\displaystyle \dfrac{1}{4}
25\displaystyle \dfrac{2}{5}
920\displaystyle \dfrac{9}{20}
12\displaystyle \dfrac{1}{2}
Answer: C
Concepts:basic probabilitysymmetry
Small hint:
A square has 44 symmetry lines
Big hint:
Each symmetry line has 88 possible points besides PP
Video solution:
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Written solution:
Note the only lines of symmetry for a square are the two diagonals and the two lines connecting opposite midpoints.
On each symmetry line there are 88 grid points other than P.P. The four lines intersect only at P,P, which is not eligible, so these give 4⋅8=324 \cdot 8 = 32 distinct choices for Q.Q.
The probability is therefore 3280=25.\dfrac{32}{80} = \dfrac{2}{5}.
Thus, the correct answer is C.
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