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| Variation on a Classic (Posted on 2005-08-23) |
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In the classic problem you are given a triangle ABC with points D on AB, E on BC, and F on AC such that |AD|=2|DB|, |BE|=2|EC|, and |CF|=2|FA|. The lines AE, BF, and CD enclose a triangle inside triangle ABC. You are to find the area of this enclosed triangle relative to that of ABC. The answer is 1/7.
What if everything is the same except |BE|=|EC| and |CF|=3|FA|. What is the area of the enclosed triangle relative to that of ABC?
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Submitted by McWorter
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Solution:
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Since a nonsingular linear transformation preserves proportion along a line segment and preserves area ratios, we may assume that A=(-1,-1), B=(0,2), and C=(3,1). Since D is a third of the way from B to A,
D=(1/3)A+(2/3)B=(1/3)(-1,-1)+(2/3)(0,2)=(-1/3,1).
Hence DC is parallel to the x-axis and all of its points have y-coordinate equal 1. Since F is one fourth of the way from A to C,
F=(3/4)A+(1/4)C=(3/4)(-1,-1)+(1/4)(3,1)=(0,-1/2).
Hence BF is coincident with the y-axis and the intersection of BF and DC is P=(0,1). Since E is the midpoint of B and C,
E=(1/2)B+(1/2)C=(1/2)(0,2)+(1/2)(3,1)=(3/2,3/2).
Hence all points on AE have their x- and y-coordinates equal. Thus the intersection of DC and AE is Q=(1,1) and the intersection of AE and BF is R=(0,0).
Thus the enclosed triangle PQR inside triangle ABC has area 1/2. The area of triangle ABC is 5, whence the area of the enclosed triangle is one tenth of the area of triangle ABC. |
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