Problem:
A point P is chosen in the interior of △ABC so that when lines are drawn through P parallel to the sides of △ABC, the resulting smaller triangles, t1,t2 and t3 in the figure, have areas 4,9 and 49, respectively. Find the area of △ABC.
Solution:
Let T denote the area of △ABC, and denote by T1,T2 and T3 the areas of t1,t2 and t3, respectively. Moreover, let c be the length of AB, and let c1,c2 and c3 be the lengths of the bases parallel to AB of t1,t2 and t3, respectively. Then, in view of the similarity of the four triangles, one has
TT1=cc1,TT2=cc2 and TT3=cc3
Moreover, since c1+c2+c3=c, it follows that
TT1+TT2+TT3=cc1+c2+c3=1
and therefore, T=(T1+T2+T3)2=(4+9+49)2=144.
