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Problem 937

AMC 12 late, AIME early

Geometry Difficulty 4.7 Find the answer HMMT November Team Round · United States · 2021

The taxicab distance between points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $|x_{2}-x_{1}|+|y_{2}-y_{1}|$ . A regular octagon is positioned in the $xy$ plane so that one of its sides has endpoints $(0,0)$ and $(1,0)$ . Let $S$ be the set of all points inside the octagon whose taxicab distance from some octagon vertex is at most $\frac{2}{3}$ . The area of $S$ can be written as $\frac{m}{n}$ , where $m, n$ are positive integers and $\operatorname{gcd}(m, n)=1$ . Find $100m+n$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Solution:

In the taxicab metric, the set of points that lie at most $d$ units away from some fixed point $P$ form a square centered at $P$ with vertices at a distance of $d$ from $P$ in directions parallel to the axes. The diagram above depicts the intersection of an octagon with eight such squares for $d=\frac{2}{3}$ centered at its vertices. (Note that since $\sqrt{2}>\frac{2}{3} \cdot 2$ , the squares centered at adjacent vertices that are diagonal from each other do not intersect.) The area of the entire shaded region is $4[ABCDEFG]=4(2([AFG]+[AYF])-[EXY])$ , which is easy to evaluate since $AFG$ , $AYF$ , and $EXY$ are all 45-45-90-degree triangles. Since $AF=\frac{2}{3}$ , $GF=\frac{\sqrt{2}}{3}$ , and $EX=\frac{1}{3\sqrt{2}}$ , the desired area is $4\left(\frac{2}{9}+\frac{4}{9}-\frac{1}{36}\right)=\frac{23}{9}$ .

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