Problem
1952 AHSME Problem 49
![[asy] unitsize(27); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A,B,C,D,E,F,X,Y,Z; A=(3,3); B=(0,0); C=(6,0); D=(4,0); E=(4,2); F=(1,1); draw(A--B--C--cycle); draw(A--D); draw(B--E); draw(C--F); X=intersectionpoint(A--D,C--F); Y=intersectionpoint(B--E,A--D); Z=intersectionpoint(B--E,C--F); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,S); label("$E$",E,NE); label("$F$",F,NW); label("$N_1$",X,NE); label("$N_2$",Y,WNW); label("$N_3$",Z,S); [/asy]](/static/aops/c1d737a4088cc4d9061c296f96c1fa7232d8d8d0.png)
In the figure, \overline{CD}, \overline{AE} and \overline{BF} are one-third of their respective sides. It follows that \overline{AN_2}: \overline{N_2N_1}: \overline{N_1D} = 3: 3: 1, and similarly for lines BE and CF. Then the area of triangle N_1N_2N_3 is:
\text{(A) } \frac {1}{10} \triangle ABC \qquad \text{(B) } \frac {1}{9} \triangle ABC \qquad \text{(C) } \frac{1}{7}\triangle ABC\qquad \text{(D) } \frac{1}{6}\triangle ABC\qquad \text{(E) } \text{none of these}
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