2019 AMC 8

Complete problem set with solutions and individual problem pages

Problem 24 Hard

In triangle \triangle ABC, point D divides side \overline{AC} so that AD:DC=1:2. Let E be the midpoint of \overline{BD} and let F be the point of intersection of line \overline{BC} and line \overline{AE}. Given that the area of \triangle ABC is 360, what is the area of \triangle EBF?

  • A.

    24

  • B.

    30

  • C.

    32

  • D.

    36

  • E.

    40

Answer:B

Solution 1

We use the line-segment ratios to infer area ratios and height ratios.

Areas:

AD:DC = 1:2 \implies AD:AC = 1:3 \implies [ABD] =\frac{[ABC]}{3} = 120.

BE:BD = 1:2 \text{ (midpoint)} \implies [ABE] = \frac{[ABD]}{2} = \frac{120}{2} = 60.

Heights:

Let h_A = height (of altitude) from \overline{BC} to A.

AD:DC = 1:2 \implies CD:CA = 2:3 \implies \text{height } h_D from \overline{BC} to D is \frac{2}{3}h_A.

BE:BD = 1:2 \text{ (midpoint)} \implies \text{height } h_E from \overline{BC} to E is \frac{1}{2} h_D = \frac{1}{2}(\frac{2}{3} h_A) = \frac{1}{3} h_A.

Conclusion:

\frac{[EBF]} {[ABF]} = \frac{[EBF]} {[EBF] + [ABE]} = \frac{[EBF]} {[EBF]+60}, and also \frac{[EBF]} {[ABF]} = \frac{h_E}{h_A} = \frac{1}{3}.

So, \frac{[EBF]} {[EBF] + 60} = \frac{1}{3}, and thus, [EBF] = \boxed{\textbf{(B) }30}

 

Solution 2

Draw X on \overline{AF} such that \overline{XD} is parallel to \overline{BC}.

Triangles BEF and EXD are similar, and since BE = ED, they are also congruent, and so XE=EF and XD=BF.

AC:AD = 3 implies \frac{AF}{AX} = 3 = \frac{FC}{XD} = \frac{FC}{BF}, so BC=BF + 3BF = 4BF, BF=\frac{BC}{4}.

Since XE=EF, AX = XE = EF, and since AX + XE + EF = AF, all of these are equal to \frac{AF}{3}, and so the altitude of triangle BEF is equal to \frac{1}{3} of the altitude of ABC.

The area of ABC is 360, so the area of\triangle EBF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}.

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