AMC 8 2018 Problem 20

AMC8 2018 年第 20 题

三角形★★★☆☆
In △ABC\triangle ABC, a point EE is on AB‾\overline{AB} with AE=1 and EB=2. Point DD is on AC‾\overline{AC} so that DE‾∥BC‾\overline{DE} \parallel \overline{BC} and point FF is on BC‾\overline{BC} so that EF‾∥AC‾\overline{EF} \parallel \overline{AC}. What is the ratio of the area of CDEFCDEF to the area of △ABC\triangle ABC?
在 △ABC\triangle ABC 中，点 EE 在边 AB‾\overline{AB} 上，满足 AE=1，EB=2。点 DD 在边 AC‾\overline{AC} 上，满足 DE‾∥BC‾\overline{DE} \parallel \overline{BC}；点 FF 在边 BC‾\overline{BC} 上，满足 EF‾∥AC‾\overline{EF} \parallel \overline{AC}。则 CDEFCDEF 的面积与 △ABC\triangle ABC 的面积之比是多少？
AMC 8 2018 Problem 20 figure
  1. A.49\displaystyle \frac{4}{9}
  2. B.12\displaystyle \frac{1}{2}
  3. C.59\displaystyle \frac{5}{9}
  4. D.35\displaystyle \frac{3}{5}
  5. E.23\displaystyle \frac{2}{3}

正确答案 · Correct Answer

A

解析 · Solution

Solution 1

By similar triangles, we have [ADE]=19[ABC]\displaystyle [ADE] = \frac{1}{9} [ABC]. Similarly, we see that [BEF]=49[ABC]\displaystyle [BEF] = \frac{4}{9} [ABC]. Using this information, we get
[ACFE]=59[ABC].[ACFE] = \frac{5}{9} [ABC].
Then, since [ADE]=19[ABC]\displaystyle [ADE] = \frac{1}{9} [ABC], it follows that the [CDEF]=49[ABC]\displaystyle [CDEF] = \frac{4}{9} [ABC]. Thus, the answer would be 49\displaystyle \frac{4}{9}. Sidenote: $[ABC]$ denotes the area of triangle ABCABC. Similarly, $[ABCD]$ denotes the area of figure ABCDABCD.

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