2018 AMC 8

Complete problem set with solutions and individual problem pages

Problem 22 Hard

Point E is the midpoint of side \overline{CD} in square ABCD, and \overline{BE} meets diagonal \overline{AC} at F. The area of quadrilateral AFED is 45. What is the area of ABCD?

  • A.

    100

  • B.

    108

  • C.

    120

  • D.

    135

  • E.

    144

Answer:B

Solution 1

We can use analytic geometry for this problem.

Let us start by giving D the coordinate (0,0), A the coordinate (0,1), and so forth. \overline{AC} and \overline{EB} can be represented by the equations y=-x+1 and y=2x-1, respectively. Solving for their intersection gives point F coordinates \left(\frac{2}{3},\frac{1}{3}\right).

Now, we can see that \triangleEFC's area is simply \frac{\frac{1}{2}\cdot\frac{1}{3}}{2} or \frac{1}{12}. This means that pentagon ABCEF's area is \frac{1}{2}+\frac{1}{12}=\frac{7}{12} of the entire square, and it follows that quadrilateral AFED's area is \frac{5}{12} of the square.

The area of the square is then \frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B) } 108}.

 

Solution 2

\triangle ABC has half the area of the square. \triangle FEC has base equal to half the square side length, and by AA Similarity with \triangle FBA, it has \frac{1}{1+2}= \frac{1}{3} the height, so has \dfrac1{12}th of the area of square(\dfrac1{2} \cdot \dfrac1{2} \cdot \dfrac1{3}). Thus, the area of the quadrilateral is 1-\frac{1}{2}-\frac{1}{12}=\frac{5}{12} of the area of the square. The area of the square is then 45\cdot\dfrac{12}{5}=\boxed{\textbf{(B) } 108}.