2019 AMC 8

Complete problem set with solutions and individual problem pages

Problem 21 Hard

What is the area of the triangle formed by the lines y=5, y=1+x, and y=1-x?

  • A.

    4

  • B.

    8

  • C.

    10

  • D.

    12

  • E.

    16

Answer:E

First, we need to find the coordinates where the graphs intersect.

We want the points x and y to be the same. Thus, we set 5=x+1, and get x=4. Plugging this into the equation, y=1-x, y=5, and y=1+x intersect at (4,5), we call this line x.

Doing the same thing, we get x=-4. Thus, y=5. Also, y=5 and y=1-x intersect at (-4,5), and we call this line y.

It's apparent the only solution to 1-x=1+x is 0. Thus, y=1. y=1-x and y=1+x intersect at (0,1), we call this line z.

Using the Shoelace Theorem we get:

\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}= So, our answer is \boxed{\textbf{(E)}\ 16.}

We might also see that the lines y and x are mirror images of each other. This is because, when rewritten, their slopes can be multiplied by -1 to get the other. As the base is horizontal, this is an isosceles triangle with base 8, as the intersection points have a distance of 8. The height is 5-1=4, so \frac{4\cdot 8}{2} = \boxed{\textbf{(E)} 16.}