1975 AMC 12 Problem 28

Attempt Problem 28 of the 1975 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1975 AMC 12 solutions, or check the answer key.

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28.

In triangle ABCABC shown in the adjoining figure, MM is the midpoint of side BC,BC, AB=12,AB=12, and AC=16.AC=16. Points EE and FF are taken on ACAC and AB,AB, respectively, and lines EFEF and AMAM intersect at G.G. If AE=2AF,AE=2AF, then EGGF\frac{EG}{GF} equals

32\frac32

43\frac43

54\frac54

65\frac65

not enough information given to solve the problem

Answer: A
Concepts:vectormedian (geometry)ratio and proportion
Difficulty rating: 2180
Small Hint:

Represent B,CB,C by vectors b,c\mathbf b,\mathbf c from AA

Big Hint:

If F=tb,F=t\mathbf b, then E=32tc;E=\frac32t\mathbf c; equate the b,c\mathbf b,\mathbf c coefficients of a point on EFEF and on AMAM

Solution:

Put AA at the origin and write the position vectors of B,CB,C as b,c.\mathbf b,\mathbf c. If F=tb,F=t\mathbf b, then AF=12tAF=12t and AE=24t,AE=24t, so E=(3t2)c.E=(\frac{3t}{2})\mathbf c. Write G=(1u)F+uE.G=(1-u)F+uE. Because GG also lies on the median AM,AM, its b\mathbf b and c\mathbf c coefficients are equal: (1u)t=32ut. (1-u)t=\frac32ut. Hence u=25.u=\frac{2}{5}. Along FE,FGFE=25FE,\frac{FG}{FE}=\frac{2}{5} and EGFE=35,\frac{EG}{FE}=\frac{3}{5}, so EGGF=32.\frac{EG}{GF}=\frac{3}{2}.

Therefore, the correct answer is A.

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