2013 AMC 10A Problem 16

Label $A = (6, 5)$, $B = (8, -3)$, $C = (9, 1)$. Reflecting over $x = 8$ replaces $x$ by $16 - x$: $A' = (10, 5)$, $B' = B$, $C' = (7, 1)$.

Key observation: the slope of $BC$ is $\frac{1 - (-3)}{9 - 8} = 4$, and the slope of $BA'$ is $\frac{5 - (-3)}{10 - 8} = 4$. So $C$ lies on segment $BA'$, and by symmetry $C'$ lies on segment $BA$.

Therefore both triangles sit inside the large triangle $ABA'$, and the union is that big triangle minus the region that belongs to neither: the small triangle at the top bounded by $AC$, $A'C'$ and the segment $AA'$.

The lines $AC$ and $A'C'$ meet on the axis $x = 8$. On $AC$, going from $(6,5)$ to $(9,1)$, $x = 8$ is two thirds of the way, so the point is $P = \left(8, 5 - \tfrac{2}{3}\cdot 4\right) = \left(8, \tfrac{7}{3}\right)$.

Areas: triangle $ABA'$ has base $AA' = 4$ and height $5 - (-3) = 8$, area $16$. Triangle $APA'$ has base $4$ and height $5 - \tfrac{7}{3} = \tfrac{8}{3}$, area $\tfrac{16}{3}$.

Union $= 16 - \tfrac{16}{3} = \tfrac{32}{3}$. The answer is $\boxed{\textbf{(E)}\ \frac{32}{3}}$.

A figure and its mirror image share the mirror line, so the union is symmetric and its shape is decided by where the edges cross that line. Always check for collinearity after reflecting; here $C$ landing on $BA'$ turns a messy union into "big triangle minus small triangle."

Union $= 2 \cdot [ABC] - [\text{overlap}]$. By shoelace $[ABC] = 8$. The overlap is symmetric about $x = 8$; its right half is triangle $BCP$ with base $BP = 3 + \tfrac{7}{3} = \tfrac{16}{3}$ along $x = 8$ and height $1$, area $\tfrac{8}{3}$. Overlap $= \tfrac{16}{3}$, union $= 16 - \tfrac{16}{3} = \tfrac{32}{3}$.

• Adding the two areas (16) without subtracting the overlap, or reflecting a vertex incorrectly (for example sending (9,1) to (7,1) but (6,5) to (8,5)).
• Computing the overlap as a triangle with vertex $C$ instead of the crossing point $P$ on the axis.