In the given figure AB and BC are adjacent sides of square ABCD; M is the midpoint of AB; N is the midpoint of BC; and AN and CM intersect at O. The ratio of the area of AOCD to the area of ABCD is m : n (GCD (m, n) = 1) then the value of 2m + 3n is equal to ________

Diagonals AC and DB are drawn. Since O is the intersection of the medians of  $\Delta ABC$ , the altitude of  $\Delta AOB$ from O is $\frac{1}{3}$ the altitude of  $\Delta ABC$  from C; i.e.  $\frac{1}{3}$ the side length, s, of the square.
Hence, area of  $\Delta AOB$  $=\frac{1}{3}$ (area of  $\Delta ABC$ ) $=\frac{1}{3}\left(\frac{1}{2}{s}^2\right)=\frac{1}{6}{s}^2$.
Similarly, area  $\Delta COB=\frac{1}{6}{s}^2$. The area of AOCD is obtained by subtracting the areas of triangles AOB and COB from that of the square.
So, area AOCD =  $s^2-\frac{1}{3}{s}^2=\frac{2}{3}{s}^2$.