2013 AMC 10A Problem 16

Question

A triangle with vertices

(6, 5),

(8, -3),

and

(9, 1)

is reflected about the line

x = 8

to create a second triangle. What is the area of the union of the two triangles?

9

\dfrac{28}{3}

10

\dfrac{31}{3}

\dfrac{32}{3}

Reflect the two non-fixed vertices across

x=8

Find the shared vertical segment created by the two overlapping triangles

The reflected triangle has vertices

(7,1)

,

(8,-3)

, and

(10,5)

.

The line through

(6,5)

and

(9,1)

is

y=-\frac43x+13

, so it meets

x=8

at

y=\frac73

. By symmetry, the union is two congruent triangles with vertical base from

(8,-3)

to

(8,\frac73)

.

That base has length

\frac73+3=\frac{16}{3}

, and each triangle has horizontal height

2

. Hence the union area is

2\left(\frac12\cdot\frac{16}{3}\cdot2\right)=\frac{32}{3}

.

Thus, E is the correct answer.

Problem 16 in other years