Problem:
In △ABC, a point E is on AB such that AE=1 and EB=2. Point D is on AC so that DE∥BC and point F is on BC so that EF∥AC. What is the ratio of the area of CDEF to the area of triangle △ABC?
Answer Choices:
A. 94
B. 21
C. 95
D. 53
E. 32
Solution:
Let the area of △ABC be equal to t. Since DE∥BC and FE∥CA, we can deduce that △ADE∼△ABC and △EFB∼△ABC. Since AE=31AB, the area of △ADE is equal to:
(31)2t=9t.
Since EB=31AB, the area of △EFB is equal to:
(32)2t=94t.
Finally, to find the area of CDEF, we take the area of △ABC=t and subtract the areas of △ADE and △EFB. This gives:
t−9t−94t=94t.
Thus, the ratio of the area of CDEF to △ABC is:
t94t=94.
Therefore, the correct answer is A.
Answer: A.
The problems on this page are the property of the MAA's American Mathematics Competitions
