2013 AMC 10A Problem 16

Call the points $A = (6, 5)$, $B = (8, -3)$, $C = (9, 1)$.

1. Reflect. The line $x = 8$ is a mirror. A point $2$ units left of it goes to $2$ units right of it, and the height stays the same.
- $A = (6, 5)$ is $2$ left, so it goes to $A' = (10, 5)$.
- $B = (8, -3)$ is on the mirror, so it stays put.
- $C = (9, 1)$ is $1$ right, so it goes to $C' = (7, 1)$.
2. Spot points on a line. Going from $B = (8, -3)$ to $C = (9, 1)$ is $1$ right and $4$ up. Going from $B$ to $A' = (10, 5)$ is $2$ right and $8$ up, the same direction twice as far. So $C$ lies on segment $BA'$. By the mirror, $C'$ lies on segment $BA$.
3. Describe the union. Draw the big triangle $A B A'$. Both small triangles fit inside it. The only part of the big triangle that is covered by neither is a small triangle at the top, between $A$, $A'$ and the point $P$ where the sides $AC$ and $A'C'$ cross.
4. Find $P$. By symmetry, $P$ is on the mirror line $x = 8$. On the side from $A = (6,5)$ to $C = (9,1)$, $x$ goes up by $3$ while $y$ goes down by $4$. To reach $x = 8$ we go $2$ of the $3$ steps, so $y$ drops by $\tfrac23 \times 4 = \tfrac83$. So $P = \left(8, 5 - \tfrac83\right) = \left(8, \tfrac73\right)$.
5. Big triangle area. Base $AA'$ runs from $x = 6$ to $x = 10$ along $y = 5$, length $4$. The height is from $y = 5$ down to $y = -3$, which is $8$. Area $= \tfrac12 \times 4 \times 8 = 16$.
6. Small top triangle area. Same base $4$. Height from $y = 5$ down to $y = \tfrac73$, which is $\tfrac83$. Area $= \tfrac12 \times 4 \times \tfrac83 = \tfrac{16}{3}$.
7. Union. $16 - \tfrac{16}{3} = \tfrac{48}{3} - \tfrac{16}{3} = \tfrac{32}{3}$.

The answer is $\boxed{\textbf{(E)}\ \frac{32}{3}}$.

A shape and its mirror image are symmetric about the mirror line, so crossing points sit on that line. After reflecting, always check whether some points line up; here $C$ landing on $BA'$ turned an odd-shaped union into "big triangle minus small triangle", which is easy to measure.

Union $=$ (area of both triangles) $-$ (overlap). Step 2 showed that $C$ is the midpoint of $BA'$. So triangle $ABC$ is exactly half of the big triangle $ABA'$ (same top corner $A$, half the base $BA'$), and its area is $8$. Two triangles give $16$. The overlap is the four-sided shape $B\,C\,P\,C'$, which the mirror line cuts into two equal triangles. The right half, $BCP$, has its side $BP$ on $x = 8$ of length $\tfrac73 - (-3) = \tfrac{16}{3}$, and $C$ is $1$ unit away from that line, so its area is $\tfrac12 \times \tfrac{16}{3} \times 1 = \tfrac83$. The overlap is $\tfrac{16}{3}$, and the union is $16 - \tfrac{16}{3} = \tfrac{32}{3}$.

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 $\boxed{\textbf{(E)}\ \frac{32}{3}}$.