smc

Question

FIGURE In the figure, CD, AE and BF are one-third of their respective sides. It follows that AN_2: N_2N_1: N_1D = 3: 3: 1, and similarly for lines BE and CF. Then the area of triangle N_1N_2N_3 is:

Accepted Answer

Let [ABC]=K. Then [ADC] = 13K, and hence [N_1DC] = 17 [ADC] = 121K. Similarly, [N_2EA]=[N_3FB] = 121K. Then [N_2N_1CE] = [ADC] - [N_1DC]-[N_2EA] = 521K, and same for the other quadrilaterals. Then [N_1N_2N_3] is just [ABC] minus all the other regions we just computed. That is, [N_1N_2N_3] = K - 3(121K) - 3(521)K = K - 67K = 17 ABC