Question

There are $81$ grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point $P$ is in the center of the square. Given that point $Q$ is randomly chosen among the other $80$ points, what is the probability that the line $PQ$ is a line of symmetry for the square?

$\displaystyle \dfrac{1}{5}$

$\displaystyle \dfrac{1}{4}$

$\displaystyle \dfrac{2}{5}$

$\displaystyle \dfrac{9}{20}$

$\displaystyle \dfrac{1}{2}$

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.

Therefore, $Q$ must be a point on any one of these four lines. Each line consists of $9$ points, for a total of $4 \cdot 9 = 36$ points.

Note that $Q$ cannot be the same as $P,$ so we have to subtract out the $4$ points which coincide with $P,$ for a total of $36 - 4 = 32$ points.

The probability is therefore $\dfrac{32}{80} = \dfrac{2}{5}.$

Thus, the correct answer is C.

2019 AMC 8 Problem 6

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Problem 6 in Other Years